Infinite Limits and Vertical Asymptotes - Maths: an open, verified university mathematics reference

Why this matters¶

A magnifying glass has a focal length of 10 cm. Put a small object $u$ centimetres in front of it. The thin-lens equation $\frac{1}{u} + \frac{1}{v} = \frac{1}{10}$ says where the lens forms the sharp image of the object: at distance

v(u) = \frac{10u}{u - 10}

(1)

behind the lens, for every $u \ne 10$ , in centimetres. (A negative $v$ means that the image is on the same side as the object: the enlarged virtual image you see when you look through the glass.) What happens as the object moves towards the focal point, $u = 10$ ?

$u$	11	10.1	10.01	9.99	9.9
$v(u)$	110	1010	$10\,010$	${-9990}$	${-990}$

From the right, the image runs away behind the lens: 10.01 cm gives an image 100 metres away. From the left, the virtual image runs away just as fast on the other side. So $v(u)$ approaches no number as $u$ approaches 10 from either side. On the page The Limit of a Function such behaviour was one of the ways a limit can fail to exist (compare its example of unbounded values, $\frac{1}{x^2}$ near 0). But “does not exist” throws away what we see in the table: on the right the values grow beyond every bound, and on the left they fall below every bound.

This page gives that behaviour a notation, $\lim_{u \to 10^{+}} v(u) = \infty$ and $\lim_{u \to 10^{-}} v(u) = -\infty$ , and a precise meaning. The notation does not make $\infty$ a number: we prove that such a limit does not exist as a real number. Then we find infinite limits of rational functions by a sign analysis, and the lines they create in the graph, the vertical asymptotes. Example 4 proves the two limits of the table, and finds how close to 10 cm the object must be for the image to be more than 10 metres away.

Limits that are infinite¶

Informally, $\lim_{x \to a^{+}} f(x) = \infty$ means that $f(x)$ is as large as we like for all $x$ close enough to $a$ and greater than $a$ . The game of the definition of one-sided limits changes in one place: instead of a band of half-width $\eps$ around a number $L$ , the challenger names a height $M$ , and the values of $f$ on the half-window must lie above it.

Two graphs side by side, each for x from 0 to 4, with a dashed vertical line at x = 2. Left, the graph of y = 1/(x − 2): to the right of the dashed line it rises steeply out of view as x approaches 2, and to the left of the line it falls steeply out of view. Right, the graph of y = 1/(x − 2) squared: on both sides of the dashed line it rises steeply out of view. — Figure 1:Left: $y = \frac{1}{x - 2}$ climbs beyond every height as $x$ approaches 2 from the right, and falls below every height as $x$ approaches 2 from the left. Right: $y = \frac{1}{(x - 2)^2}$ climbs beyond every height from both sides. The dashed line $x = 2$ is a vertical asymptote of both graphs. A drawing shows finitely many points; Proposition 2 proves what it suggests.

Definition 1 (Infinite limits)

Let $a$ be a real number.

(a) Let $f$ be a function that is defined at every point of an open interval $(a, a + r)$ , for some $r > 0$ . We say that $f(x)$ tends to $\infty$ as $x$ approaches $a$ from the right, and write

\lim_{x \to a^{+}} f(x) = \infty,

(2)

if, whatever height $M > 0$ is named, $f(x) > M$ for all $x > a$ that are close enough to $a$ . We say that $f(x)$ tends to $-\infty$ as $x \to a^{+}$ , and write $\lim_{x \to a^{+}} f(x) = -\infty$ , if, whatever $M > 0$ is named, $f(x) < -M$ for all $x > a$ that are close enough to $a$ .

(b) Let $f$ be a function that is defined at every point of an open interval $(a - r, a)$ , for some $r > 0$ . The statements $\lim_{x \to a^{-}} f(x) = \infty$ and $\lim_{x \to a^{-}} f(x) = -\infty$ say the same for all $x < a$ that are close enough to $a$ .

(c) Let $f$ be a function that is defined at every point of $(a - r, a)$ and of $(a, a + r)$ , for some $r > 0$ . We write $\lim_{x \to a} f(x) = \infty$ if both $\lim_{x \to a^{-}} f(x) = \infty$ and $\lim_{x \to a^{+}} f(x) = \infty$ , and $\lim_{x \to a} f(x) = -\infty$ if both one-sided limits are $-\infty$ .

These are infinite limits. We also write $f(x) \to \infty$ as $x \to a^{+}$ , and so on.

“Close enough” means the same as in the definition of one-sided limits: there is a half-window $a < x < a + \delta$ (or $a - \delta < x < a$ ) on which it holds. The rigorous track states this precisely, in the dropdown below; the proofs on this page use that version.

Definition 2 (Infinite limits, precisely (rigorous track))

Let $a$ be a real number.

(a) Let $f$ be as in part (a) of Definition 1. Then $\lim_{x \to a^{+}} f(x) = \infty$ means: for every $M > 0$ there is a $\delta > 0$ such that $f$ is defined at every $x$ with $a < x < a + \delta$ , and

\begin{aligned} &a < x < a + \delta \\ &\quad\text{implies}\quad f(x) > M . \end{aligned}

(3)

And $\lim_{x \to a^{+}} f(x) = -\infty$ means the same with $f(x) < -M$ in place of $f(x) > M$ .

(b) Let $f$ be as in part (b) of Definition 1. The statements $\lim_{x \to a^{-}} f(x) = \infty$ and $\lim_{x \to a^{-}} f(x) = -\infty$ mean the same as in (a), with the half-window $a - \delta < x < a$ in place of $a < x < a + \delta$ .

Part (c) of Definition 1 is defined through (a) and (b), so it needs no precise version of its own.

In words. A challenger names a height $M > 0$ ; you reply with a half-window of width $\delta$ next to $a$ ; you win the round if $f$ is defined at every $x$ of the half-window and every such $f(x)$ is above $M$ . The limit is $\infty$ when you can win every round. Three points to notice:

$\delta$ is chosen after $M$ , so it may depend on $M$ (and on $a$ ), but not on $x$ . A higher $M$ usually needs a narrower half-window. If a $\delta$ wins a round, so does every smaller $\delta > 0$ , and it wins every lower height too.
Neither half-window contains $a$ . The value $f(a)$ , and whether it exists, plays no part.
“ $= \infty$ ” is notation, not an equation between numbers: $\infty$ is not a real number, and there is no band around it. It says how $f(x)$ fails to approach a number. Proposition 1 below proves that it does fail.

Example. $\lim_{x \to 0^{+}} \frac{1}{x} = \infty$ . The function $\frac{1}{x}$ is defined at every $x \ne 0$ . Given $M > 0$ , take $\delta = \frac{1}{M}$ , which is positive by property 6 of the order rules. If $0 < x < \frac{1}{M}$ , then by the same property 6 ( $0 < u < v$ implies $\frac{1}{v} < \frac{1}{u}$ , with $u = x$ and $v = \frac{1}{M}$ ), $M = \frac{1}{1/M} < \frac{1}{x}$ .

Non-example. $\lim_{x \to 0^{-}} \frac{1}{x}$ is not $\infty$ . Take $M = 1$ . For $x < 0$ , $\frac{1}{x}$ has the sign of $x$ (part (c) of the sign rules), so $\frac{1}{x} < 0 < 1$ . So every point of every half-window $-\delta < x < 0$ fails the round $M = 1$ , and no $\delta$ wins it. (From the left, $\frac{1}{x}$ tends to $-\infty$ : Proposition 2 below.)

An infinite limit at $a$ shows in the graph of $f$ as a vertical line that the graph climbs or falls along without bound, as the line $x = 2$ in Figure 1.

In words. On at least one side of the line $x = a$ , the graph of $f$ climbs beyond every height or falls below every height as $x$ approaches $a$ . One side is enough, and the two sides may go in different directions. As with every limit, the value $f(a)$ plays no part: $f$ may be defined at $a$ and still have the vertical asymptote $x = a$ . For example, $F(x) = \frac{1}{x} + 1$ for $x > 0$ and $F(x) = 0$ for $x \le 0$ is defined at 0, and $\lim_{x \to 0^{+}} F(x) = \infty$ : for $x > 0$ , adding $\frac{1}{x}$ to $0 < 1$ gives $\frac{1}{x} < F(x)$ (property 3 of the order rules), so by transitivity (property 2) the $\delta = \frac{1}{M}$ of the example above wins every round $M$ for $F$ too.

Example. The line $x = 0$ is a vertical asymptote of the graph of $\frac{1}{x}$ , by the example after Definition 2.

Non-example. Let $g(x) = \dfrac{x^2 - 1}{x - 1}$ for $x \ne 1$ . Although $g$ is undefined at 1, the line $x = 1$ is not a vertical asymptote of its graph. For $x \ne 1$ we may cancel the non-zero factor $x - 1$ , so $g(x) = x + 1$ . If $0 < x < 2$ , adding 1 keeps both inequalities (property 3 of the order rules): $1 < g(x) < 3$ for every $x \ne 1$ in $(0, 2)$ . Take $M = 3$ . Every half-window $1 < x < 1 + \delta$ contains the point $x = 1 + \frac12 \min(\delta, 1)$ , which lies in $(1, 2)$ , so $g(x) < 3 = M$ there: no $\delta$ wins the round $M = 3$ , and $\lim_{x \to 1^{+}} g(x) \ne \infty$ . At the same point $g(x) > 1 > -3$ , so $\lim_{x \to 1^{+}} g(x) \ne -\infty$ either, and the point $x = 1 - \frac12 \min(\delta, 1)$ of the left half-window does the same from the left. A function that stays between two numbers near $a$ has no vertical asymptote at $a$ .

Main results¶

An infinite limit is not a real limit¶

If $f(x)$ is above every height near $a$ , it cannot also stay within 1 of a fixed number $L$ , because $L + 1$ is one of the heights.

Proposition 1 (An infinite limit is not a real limit)

Let $a$ be a real number.

(a) Let $f$ be as in part (a) of Definition 1. If $\lim_{x \to a^{+}} f(x) = \infty$ or $\lim_{x \to a^{+}} f(x) = -\infty$ , then there is no real number $L$ with $\lim_{x \to a^{+}} f(x) = L$ : the right-hand limit of $f$ at $a$ does not exist. The same holds for left-hand limits, with $f$ as in part (b) of Definition 1.

(b) Let $f$ be defined at every point of $(a - r, a)$ and of $(a, a + r)$ , for some $r > 0$ . If at least one of the one-sided limits of $f$ at $a$ is $\infty$ or $-\infty$ , then $\lim_{x \to a} f(x)$ does not exist. In particular, it does not exist if $\lim_{x \to a} f(x) = \infty$ or $\lim_{x \to a} f(x) = -\infty$ .

Proof

For (a) we argue by contradiction: a real limit $L$ keeps $f(x)$ below $L + 1$ on a half-window, an infinite limit pushes it above $\abs{L} + 1 \ge L + 1$ on another, and one point lies in both. Part (b) then follows from the corollary on one-sided limits.

(a) First let $\lim_{x \to a^{+}} f(x) = \infty$ , and suppose that also $\lim_{x \to a^{+}} f(x) = L$ for a real number $L$ . By part (a) of Definition 1, with $\eps = 1$ , there is a $\delta_1 > 0$ such that $f$ is defined and $\abs{f(x) - L} < 1$ at every $x$ with $a < x < a + \delta_1$ . By part (a) of the proposition on distance inequalities, with $f(x)$ , $L$ and 1 in place of $x$ , $a$ and $\delta$ , every such $x$ has $L - 1 < f(x) < L + 1$ .

Let $M = \abs{L} + 1$ . Since $\abs{L} \ge 0$ (property 1 of the absolute value), adding 1 gives $M \ge 1$ (property 3 of the order rules), and $1 > 0$ (part (c) of the sign rules), so $M > 0$ by transitivity (property 2). By part (a) of Definition 2, there is a $\delta_2 > 0$ such that $f(x) > M$ at every $x$ with $a < x < a + \delta_2$ .

Let $\delta = \min(\delta_1, \delta_2)$ , which is positive, and $x = a + \frac{\delta}{2}$ . Multiplying $0 < \delta$ by the positive number $\frac12$ keeps the inequality (property 5(a)), so $0 < \frac{\delta}{2}$ ; adding $\frac{\delta}{2}$ to both sides (property 3) gives $\frac{\delta}{2} < \delta$ . Adding $a$ (property 3): $a < x < a + \delta$ . Since $\delta \le \delta_1$ , adding $a$ gives $a + \delta \le a + \delta_1$ , so $a < x < a + \delta_1$ (property 2); in the same way $a < x < a + \delta_2$ . So $x$ lies in both half-windows, and

\begin{aligned} f(x) &< L + 1 \le \abs{L} + 1 \\ &= M < f(x) . \end{aligned}

(4)

Here $L \le \abs{L}$ is property 3 of the absolute value, and adding 1 keeps it (property 3 of the order rules). By transitivity (property 2), $f(x) < f(x)$ , which trichotomy (property 1) forbids. So no such $L$ exists.

If instead $\lim_{x \to a^{+}} f(x) = -\infty$ , take the same $M$ and a $\delta_2 > 0$ with $f(x) < -M$ for $a < x < a + \delta_2$ . Property 3 of the absolute value also gives $-\abs{L} \le L$ , and adding -1 keeps it, so at the same point $x$

\begin{aligned} -M &= -\abs{L} - 1 \le L - 1 \\ &< f(x) < -M , \end{aligned}

(5)

and again $-M < -M$ , which is impossible.

For left-hand limits, use the point $x = a - \frac{\delta}{2}$ of the half-windows $a - \delta_1 < x < a$ and $a - \delta_2 < x < a$ . Multiplying $0 < \frac{\delta}{2} < \delta$ by the negative number ${-1}$ reverses both inequalities (property 5(c)), and adding $a$ keeps them: $a - \delta < x < a$ . Multiplying $\delta \le \delta_1$ by ${-1}$ and adding $a$ gives $a - \delta_1 \le a - \delta$ , so $x$ lies in the half-window of $\delta_1$ , and in the same way in that of $\delta_2$ . The rest is unchanged.

(b) Since $r > 0$ , adding $a$ and $a - r$ gives $a - r < a < a + r$ (property 3), so $(a - r, a + r)$ is an open interval containing $a$ , and $f$ is defined at every point of it except possibly at $a$ : $f$ and $a$ are as in the definition of the limit. By (a), the one-sided limit that is $\infty$ or $-\infty$ does not exist. By part (b) of the corollary on differing one-sided limits, $\lim_{x \to a} f(x)$ does not exist. If $\lim_{x \to a} f(x) = \infty$ or $-\infty$ , then by part (c) of Definition 1 both one-sided limits are, so this applies.

In words. “ $\lim_{x \to a} f(x) = \infty$ ” is a precise way of saying that the limit does not exist: it records the reason. The converse is false. A limit can fail to exist without being $\infty$ or $-\infty$ : at a jump, such as $\frac{\abs{x}}{x}$ at 0, or when the two sides go to different infinities, as $\frac{1}{x}$ does at 0 (Proposition 2 below), so that $\lim_{x \to 0} \frac{1}{x}$ is neither $\infty$ nor $-\infty$ , by part (c) of Definition 1.

Reciprocal powers¶

The simplest infinite limits are those of $\frac{1}{(x - a)^n}$ . Their sign depends on the side and on whether $n$ is even or odd (Figure 1).

Proof

Let $t = \abs{x - a}$ be the distance from $x$ to $a$ . For $0 < t \le 1$ , the power $t^n$ is at most $t$ , so $\frac{1}{t^n} \ge \frac{1}{t}$ , which is above $M$ when $t < \frac{1}{M}$ : the half-windows of width $\delta = \min\bigl(1, \frac{1}{M}\bigr)$ win the round $M$ . The sign of $(x - a)^n$ then decides between $\infty$ and $-\infty$ .

The domain. For $x \ne a$ , each of the $n$ factors of $(x - a)^n$ is non-zero, so the product is non-zero, by part (a) of the sign rules applied to one factor after another. So $g$ is defined at every $x \ne a$ , in particular at every point of $(a - 1, a)$ and of $(a, a + 1)$ .

Step 1: powers of a number in $(0, 1]$ . Let $0 < t \le 1$ . We show that $0 < t^k \le t$ for every integer $k \ge 1$ . For $k = 1$ this says $0 < t \le t$ . Suppose it holds for some $k$ . Then $t^{k+1} = t^k \cdot t$ is a product of two positive numbers, so it is positive (part (c) of the sign rules); and multiplying $t \le 1$ by the positive number $t^k$ gives $t^{k+1} \le t^k$ (property 5(b) of the order rules), so $t^{k+1} \le t^k \le t$ by transitivity (property 2). So the claim passes from each $k$ to $k + 1$ , and therefore holds for every $k \ge 1$ . (If it failed for some $k$ , the set of integers $k \ge 1$ at which it fails would be a non-empty set of positive integers, so it would have a smallest element $k$ , by the well-ordering principle. The claim holds for 1, so $k \ge 2$ . Then $k - 1 \ge 1$ is smaller than the smallest failure, so the claim holds for $k - 1$ , and the step above gives it for $k$ : a contradiction.)

Step 2: the size and the sign of $(x - a)^n$ . Let $x \ne a$ and $t = \abs{x - a}$ . By property 4 of the absolute value, $\abs{uv} = \abs{u}\,\abs{v}$ , applied to one factor after another, $\abs{(x - a)^n} = t^n$ . If $x > a$ , each factor $x - a$ is positive (part (b) of the sign rules), and so is the product. If $x < a$ , all $n$ factors are negative, and by part (d) of the sign rules the product is positive if $n$ is even and negative if $n$ is odd. By the definition of the absolute value, a positive number equals its absolute value and a negative one equals minus its absolute value. So $(x - a)^n = t^n$ if $x > a$ , or if $x < a$ and $n$ is even, and $(x - a)^n = -t^n$ if $x < a$ and $n$ is odd.

Step 3: the height $M$ . Let $M > 0$ and $\delta = \min\bigl(1, \frac{1}{M}\bigr)$ , which is positive, because $\frac{1}{M} > 0$ (property 6 of the order rules). Let $0 < t < \delta$ . Then $t < 1$ , so step 1 gives $0 < t^n \le t$ ; and $t < \frac{1}{M}$ , so $0 < t^n < \frac{1}{M}$ (property 2). By property 6 ( $0 < u < v$ implies $\frac{1}{v} < \frac{1}{u}$ ), with $u = t^n$ and $v = \frac{1}{M}$ ,

M = \frac{1}{1/M} < \frac{1}{t^n} .

(6)

(a) Let $a < x < a + \delta$ . Adding $-a$ (property 3) gives $0 < x - a < \delta$ , so $t = x - a$ satisfies $0 < t < \delta$ . By step 2, $(x - a)^n = t^n$ , so $g(x)$ is defined and, by step 3, $g(x) = \frac{1}{t^n} > M$ . So $\delta$ wins the round $M$ , and since $M > 0$ was arbitrary, $\lim_{x \to a^{+}} g(x) = \infty$ by part (a) of Definition 2.

(b) and (c). Let $a - \delta < x < a$ . Then $t = \abs{x - a} = a - x$ , because $x - a < 0$ (the definition of the absolute value). Adding $-x$ to $x < a$ gives $0 < a - x$ , and adding $\delta - x$ to $a - \delta < x$ gives $a - x < \delta$ (property 3), so $0 < t < \delta$ , and step 3 gives $\frac{1}{t^n} > M$ .

If $n$ is even, $(x - a)^n = t^n$ by step 2, so $g(x) = \frac{1}{t^n} > M$ . So $\lim_{x \to a^{-}} g(x) = \infty$ , by part (b) of Definition 2. With (a), both one-sided limits are $\infty$ , and $g$ is defined on $(a - 1, a)$ and on $(a, a + 1)$ , so $\lim_{x \to a} g(x) = \infty$ by part (c) of Definition 1.
If $n$ is odd, $(x - a)^n = -t^n$ by step 2, so $g(x) = -\frac{1}{t^n}$ . Multiplying $M < \frac{1}{t^n}$ by the negative number ${-1}$ reverses it (property 5(c)): $g(x) = -\frac{1}{t^n} < -M$ . So $\lim_{x \to a^{-}} g(x) = -\infty$ .

So for every $n \ge 1$ the line $x = a$ is a vertical asymptote of the graph of $\frac{1}{(x - a)^n}$ . For odd $n$ the one-sided limits are $\infty$ and $-\infty$ , and $\lim_{x \to a} \frac{1}{(x - a)^n}$ is neither, by part (c) of Definition 1.

Sign analysis: a factor that stays away from 0¶

Near a zero $a$ of its denominator, after cancelling common factors, a rational function is a reciprocal power $\frac{1}{(x - a)^k}$ times a factor $h(x)$ that does not vanish at $a$ . If $h$ keeps one sign and stays a fixed distance away from 0 near $a$ , it cannot stop the reciprocal power from growing; it can only flip its sign.

Proposition 3 (A factor that stays away from 0)

Let $a$ be a real number and $r > 0$ , and let $f$ , $g$ and $h$ be functions that are defined at every point of $(a, a + r)$ , with $f(x) = h(x)\, g(x)$ for every $x$ in $(a, a + r)$ . Suppose that $\lim_{x \to a^{+}} g(x)$ is $\infty$ or $-\infty$ .

(a) If there is a number $m > 0$ with $h(x) \ge m$ for every $x$ in $(a, a + r)$ , then $\lim_{x \to a^{+}} f(x)$ is $\infty$ if $\lim_{x \to a^{+}} g(x) = \infty$ , and $-\infty$ if $\lim_{x \to a^{+}} g(x) = -\infty$ .

(b) If there is a number $m > 0$ with $h(x) \le -m$ for every $x$ in $(a, a + r)$ , then $\lim_{x \to a^{+}} f(x)$ is $-\infty$ if $\lim_{x \to a^{+}} g(x) = \infty$ , and $\infty$ if $\lim_{x \to a^{+}} g(x) = -\infty$ .

(c) If $\lim_{x \to a^{+}} h(x) = L$ for a real number $L \ne 0$ , then the conclusion of (a) holds if $L > 0$ , and that of (b) if $L < 0$ .

The same holds for left-hand limits, with $(a - r, a)$ in place of $(a, a + r)$ .

Proof

To push $f(x) = h(x)\,g(x)$ above a height $M$ , we ask $g$ for the height $\frac{M}{m}$ : then $h(x)\,g(x) \ge m\,g(x) > M$ . Changing the sign of $h$ or of $g$ changes the sign of the conclusion, and a limit $L \ne 0$ of $h$ keeps $h$ a fixed distance from 0.

Changing sign. For a function $F$ defined on $(a, a + r)$ , $\lim_{x \to a^{+}} F(x) = -\infty$ exactly when $\lim_{x \to a^{+}} \bigl(-F(x)\bigr) = \infty$ : multiplying by the negative number ${-1}$ reverses an inequality (property 5(c) of the order rules), so $F(x) < -M$ exactly when $-F(x) > M$ , and the same $\delta$ wins the same rounds in Definition 2. $(\ast)$

(a) First let $\lim_{x \to a^{+}} g(x) = \infty$ , and let $M > 0$ . The number $\frac{M}{m}$ is a quotient of two positive numbers, so it is positive (part (c) of the sign rules). By part (a) of Definition 2, with the height $\frac{M}{m}$ , there is a $\delta_1 > 0$ such that $g(x) > \frac{M}{m}$ at every $x$ with $a < x < a + \delta_1$ . Let $\delta = \min(\delta_1, r)$ , which is positive, and let $a < x < a + \delta$ . Since $\delta \le \delta_1$ and $\delta \le r$ , adding $a$ (property 3) and transitivity (property 2) give $x < a + \delta_1$ and $x < a + r$ . So $x$ lies in $(a, a + r)$ : $f(x)$ is defined, $f(x) = h(x)\,g(x)$ and $h(x) \ge m$ ; and $g(x) > \frac{M}{m} > 0$ . Multiplying $h(x) \ge m$ by the positive number $g(x)$ gives $h(x)\,g(x) \ge m\,g(x)$ (property 5(b)), and multiplying $g(x) > \frac{M}{m}$ by the positive number $m$ gives $m\,g(x) > M$ (property 5(a)). By property 2,

f(x) = h(x)\,g(x) \ge m\,g(x) > M ,

(7)

so $f(x) > M$ . So $\delta$ wins the round $M$ , and $\lim_{x \to a^{+}} f(x) = \infty$ .

Now let $\lim_{x \to a^{+}} g(x) = -\infty$ . By $(\ast)$ , $-g$ tends to $\infty$ , so by the case just proved $h \cdot (-g) = -f$ tends to $\infty$ , and by $(\ast)$ again $f$ tends to $-\infty$ .

(b) If $h(x) \le -m$ , then $-h(x) \ge m$ (property 5(c)), and $f = (-h) \cdot (-g)$ . If $g$ tends to $\infty$ , then $-g$ tends to $-\infty$ by $(\ast)$ , and (a), applied to $-h$ and $-g$ , gives $\lim_{x \to a^{+}} f(x) = -\infty$ . If $g$ tends to $-\infty$ , then $-g$ tends to $\infty$ , and (a) gives $\lim_{x \to a^{+}} f(x) = \infty$ .

(c) Let $L > 0$ . Then $\frac{L}{2} > 0$ (property 5(a)). By part (a) of Definition 1, with $\eps = \frac{L}{2}$ , there is a $\delta_0 > 0$ such that $h$ is defined and $\abs{h(x) - L} < \frac{L}{2}$ at every $x$ with $a < x < a + \delta_0$ . By part (a) of the proposition on distance inequalities, each such $x$ has $h(x) > L - \frac{L}{2} = \frac{L}{2}$ . Let $r' = \min(r, \delta_0)$ , which is positive. Every $x$ in $(a, a + r')$ lies in $(a, a + r)$ and in the half-window of $\delta_0$ , as in (a), so $h(x) \ge \frac{L}{2}$ and $f(x) = h(x)\,g(x)$ there. So (a) applies with $r'$ in place of $r$ and $m = \frac{L}{2}$ . If $L < 0$ , take $\eps = -\frac{L}{2}$ , which is positive (property 5(c)); now each $x$ of the half-window has $h(x) < L + \bigl(-\frac{L}{2}\bigr) = \frac{L}{2}$ , so $h(x) \le -m$ with $m = -\frac{L}{2} > 0$ , and (b) applies.

For left-hand limits, use the half-windows $a - \delta < x < a$ throughout. Multiplying $\delta \le \delta_1$ and $\delta \le r$ by ${-1}$ reverses them (property 5(c)), and adding $a$ keeps them, so $a - \delta_1 \le a - \delta < x$ and $a - r \le a - \delta < x$ : $x$ lies in $(a - r, a)$ and in the half-window of $\delta_1$ . The rest is unchanged.

In words: how to find the one-sided limits of a rational function at a zero $a$ of its denominator.

Factor the numerator and the denominator. If the numerator is also 0 at $a$ , cancel the common factor $x - a$ first: for $x \ne a$ the two expressions are the same number. What is left may have a finite limit at $a$ (Example 3).
Sign. By the proposition on the sign of a factored rational function, $f$ has one sign on the interval just to the right of $a$ and one just to the left (up to the next number of a linear factor). These signs tell you which of $\infty$ and $-\infty$ to expect on each side.
Split. On a window around $a$ , write $f(x) = h(x) \cdot \frac{1}{(x - a)^k}$ , where $k$ is the number of factors $x - a$ in the denominator. This is algebra at each point $x \ne a$ of the window, not a statement about limits.
Bound. Choose the window so small that it contains no other number of a linear factor, and find an $m > 0$ with $h(x) \ge m$ (or $h(x) \le -m$ ) at every point of it, with the order rules.
Conclude with Proposition 2, for $\frac{1}{(x - a)^k}$ , and Proposition 3.

Worked examples¶

One-sided limits at a zero of the denominator¶

Example 1 (The one-sided limits of

\frac{x + 1}{x - 2}

at 2)

Let $f(x) = \dfrac{x + 1}{x - 2}$ for $x \ne 2$ . Find both one-sided limits of $f$ at 2, decide whether the line $x = 2$ is a vertical asymptote, and whether $\lim_{x \to 2} f(x)$ exists.

Domain. The denominator $x - 2$ is 0 only at $x = 2$ (part (b) of the sign rules), so $f$ is defined at every point of $(1, 2)$ and of $(2, 3)$ .
Sign. By Proposition 4 (with $K = 1$ and the numbers ${-1}$ and 2), $f$ has one sign on $(-1, 2)$ and one on $(2, \infty)$ . At $x = 3$ both factors $x + 1$ and $x - 2$ are positive, so $f > 0$ on $(2, \infty)$ ; at $x = 0$ only $x - 2$ is negative, so $f < 0$ on $(-1, 2)$ . We expect $\infty$ on the right and $-\infty$ on the left.
Split. For $x \ne 2$ , $f(x) = h(x)\,g(x)$ with $h(x) = x + 1$ and $g(x) = \frac{1}{x - 2}$ .
Bound. If $1 < x < 3$ , adding 1 keeps both inequalities (property 3 of the order rules): $2 < x + 1 < 4$ . So $h(x) \ge 2$ at every point of $(1, 2)$ and of $(2, 3)$ ; take $m = 2$ .
From the right. By part (a) of the proposition on reciprocal powers, with $a = 2$ and $n = 1$ , $\lim_{x \to 2^{+}} g(x) = \infty$ . By part (a) of Proposition 3, on $(2, 3)$ , $\lim_{x \to 2^{+}} f(x) = \infty$ .
From the left. By part (c) of Proposition 2 ( $n = 1$ is odd), $\lim_{x \to 2^{-}} g(x) = -\infty$ . By the left-hand version of part (a) of Proposition 3, on $(1, 2)$ , $\lim_{x \to 2^{-}} f(x) = -\infty$ .
Conclusions. The line $x = 2$ is a vertical asymptote (Definition 3). By part (b) of Proposition 1, $\lim_{x \to 2} f(x)$ does not exist; and it is neither $\infty$ nor $-\infty$ , because the two one-sided limits differ (part (c) of Definition 1).

\boxed{ \begin{aligned} \lim_{x \to 2^{-}} \frac{x + 1}{x - 2} &= -\infty, \\ \lim_{x \to 2^{+}} \frac{x + 1}{x - 2} &= \infty \end{aligned} }

(8)

Check. $f(2.001) = \frac{3.001}{0.001} = 3001$ and $f(1.999) = \frac{2.999}{-0.001} = -2999$ : large, with the signs of step 2. ✓ For $M = 1000$ , the proof of Proposition 3 asks $g$ for the height $\frac{M}{m} = 500$ , and the proof of Proposition 2 answers with $\delta = \min\bigl(1, \frac{1}{500}\bigr) = 0.002$ . The point $x = 2.0019$ of that half-window gives $f(x) = \frac{3.0019}{0.0019} \approx 1579.9 > 1000$ . ✓

Example 2 (A sign analysis at two zeros of the denominator)

Let $f(x) = \dfrac{x - 3}{x^3 - x^2}$ . Find the one-sided limits of $f$ at each zero of its denominator.

Factor. $x^3 - x^2 = x \cdot x \cdot (x - 1)$ , which is 0 exactly when $x = 0$ or $x = 1$ (part (a) of the sign rules). The numerator $x - 3$ is not 0 at either. So $f$ is defined at every $x \ne 0, 1$ .
Sign. The numbers 0, 1 and 3 of the linear factors cut $\R$ into four open intervals, and on each of them $f$ has one sign (Proposition 4, $K = 1$ ). We count the negative factors among $x - 3$ , $x$ , $x$ and $x - 1$ at one point of each:
Interval point negative factors $f$
$(-\infty, 0)$ ${-1}$ 4 $+$
$(0, 1)$ $\frac12$ 2 $+$
$(1, 3)$ 2 1 $-$
$(3, \infty)$ 4 0 $+$
So we expect $\infty$ on both sides of 0, $\infty$ on the left of 1, and $-\infty$ on the right of 1.
At 0: split and bound. For $x \ne 0, 1$ , $f(x) = h_0(x) \cdot \frac{1}{x^2}$ with $h_0(x) = \frac{x - 3}{x - 1} = \frac{3 - x}{1 - x}$ . Let $-\frac12 < x < \frac12$ . Multiplying by ${-1}$ reverses both inequalities (property 5(c) of the order rules), and adding 3, or 1, keeps them (property 3): $\frac52 < 3 - x < \frac72$ and $\frac12 < 1 - x < \frac32$ . Since $0 < \frac12 < 1 - x$ , the number $1 - x$ is positive (transitivity, property 2), and property 6 gives $\frac{1}{1 - x} > \frac23$ . Multiplying $3 - x > \frac52$ by the positive number $\frac{1}{1 - x}$ , and then $\frac{1}{1 - x} > \frac23$ by the positive number $\frac52$ , keeps both (property 5(a)):
$\begin{aligned} h_0(x) &= (3 - x) \cdot \frac{1}{1 - x} \\ &> \frac52 \cdot \frac{1}{1 - x} \\ &> \frac52 \cdot \frac23 = \frac53 . \end{aligned}$
(9)
So $h_0(x) \ge \frac53$ on $\bigl(-\frac12, 0\bigr)$ and on $\bigl(0, \frac12\bigr)$ .
At 0: conclude. By part (b) of the proposition on reciprocal powers ( $a = 0$ , $n = 2$ ), $\frac{1}{x^2}$ tends to $\infty$ from both sides. By part (a) of Proposition 3 and its left-hand version, $f(x) \to \infty$ as $x \to 0^{+}$ and as $x \to 0^{-}$ , so $\lim_{x \to 0} f(x) = \infty$ (part (c) of Definition 1).
At 1: split and bound. For $x \ne 0, 1$ , $f(x) = h_1(x) \cdot \frac{1}{x - 1}$ with $h_1(x) = \frac{x - 3}{x^2}$ . Let $\frac12 < x < \frac32$ . Adding -3 gives $x - 3 < -\frac32$ (property 3). Since $x > 0$ , multiplying $x < \frac32$ by the positive number $x$ , and then by the positive number $\frac32$ , gives $x^2 < \frac32 x < \frac94$ (property 5(a)), and $x^2 > 0$ (part (c) of the sign rules); so $\frac{1}{x^2} > \frac49$ by property 6. Multiplying $x - 3 < -\frac32$ by the positive number $\frac{1}{x^2}$ keeps the inequality (property 5(a)), and multiplying $\frac{1}{x^2} > \frac49$ by the negative number $-\frac32$ reverses it (property 5(c)):
$\begin{aligned} h_1(x) &= (x - 3) \cdot \frac{1}{x^2} \\ &< -\frac32 \cdot \frac{1}{x^2} \\ &< -\frac32 \cdot \frac49 = -\frac23 . \end{aligned}$
(10)
So $h_1(x) \le -\frac23$ on $\bigl(\frac12, 1\bigr)$ and on $\bigl(1, \frac32\bigr)$ .
At 1: conclude. By parts (a) and (c) of Proposition 2 ( $a = 1$ , $n = 1$ ), $\frac{1}{x - 1}$ tends to $\infty$ from the right and to $-\infty$ from the left. By part (b) of Proposition 3, which flips the sign, $f(x) \to -\infty$ as $x \to 1^{+}$ and $f(x) \to \infty$ as $x \to 1^{-}$ .

Both lines $x = 0$ and $x = 1$ are vertical asymptotes, and the signs agree with step 2.

\boxed{ \begin{aligned} \lim_{x \to 0} f(x) &= \infty, \\ \lim_{x \to 1^{-}} f(x) &= \infty, \\ \lim_{x \to 1^{+}} f(x) &= -\infty \end{aligned} }

(11)

Check. $f(0.01) = \frac{-2.99}{0.000001 - 0.0001} \approx 30\,202$ , $f(0.99) = \frac{-2.01}{0.970299 - 0.9801} \approx 205.08$ and $f(1.01) = \frac{-1.99}{1.030301 - 1.0201} \approx -195.08$ . ✓ The bound of step 5 at $x = 1.01$ : $h_1(1.01) = \frac{-1.99}{1.0201} \approx -1.95 \le -\frac23$ . ✓

A zero of the denominator that is not an asymptote¶

Example 3 (A hole and an asymptote:

\frac{x^2 - 1}{x^2 - 3x + 2}

)

Let $f(x) = \dfrac{x^2 - 1}{x^2 - 3x + 2}$ . Which zeros $a$ of the denominator give a vertical asymptote $x = a$ of its graph?

Factor. $x^2 - 1 = (x - 1)(x + 1)$ and $x^2 - 3x + 2 = (x - 1)(x - 2)$ . The denominator is 0 exactly at $x = 1$ and $x = 2$ (part (a) of the sign rules), so $f$ is defined at every $x \ne 1, 2$ . We examine these two zeros of the denominator.
Cancel. At every $x \ne 1, 2$ the factor $x - 1$ is not 0, so we may cancel it:
$f(x) = \frac{x + 1}{x - 2} \qquad (x \ne 1, 2).$
(12)
This is an equation between two numbers at each such $x$ , not a statement about limits.
At 1: a finite limit. $f$ is defined at every point of the open interval $(0, 2)$ except at 1, so the definition of the limit applies. We show $\lim_{x \to 1} f(x) = -2$ . For $x \ne 1, 2$ , by step 2,
$\begin{aligned} f(x) + 2 &= \frac{x + 1 + 2(x - 2)}{x - 2} \\ &= \frac{3(x - 1)}{x - 2}, \end{aligned}$
(13)
so $\abs{f(x) + 2} = \frac{3\abs{x - 1}}{\abs{x - 2}}$ (property 4 of the absolute value). If $\abs{x - 1} < \frac12$ , then $\frac12 < x < \frac32$ (part (a) of the proposition on distance inequalities), so $x - 2 < -\frac12$ (adding -2, property 3). Multiplying by ${-1}$ reverses this (property 5(c) of the order rules): $2 - x > \frac12$ . So $\abs{x - 2} = 2 - x > \frac12$ (the definition of the absolute value); by property 6 of the order rules, $\frac{1}{\abs{x - 2}} < 2$ . Let $\eps > 0$ , $\delta = \min\bigl(\frac12, \frac{\eps}{6}\bigr)$ , and $0 < \abs{x - 1} < \delta$ . Then $x \ne 1$ and $\frac12 < x < \frac32$ , so $x \ne 2$ and $f(x)$ is defined. Multiplying $\frac{1}{\abs{x - 2}} < 2$ by the positive number $3\abs{x - 1}$ keeps it (property 5(a)); and multiplying $\abs{x - 1} < \frac{\eps}{6}$ by the positive number 6 keeps it too (property 5(a)):
$\begin{aligned} \abs{f(x) - (-2)} &< 6\abs{x - 1} \\ &< 6 \cdot \frac{\eps}{6} = \eps . \end{aligned}$
(14)
So $\lim_{x \to 1} f(x) = -2$ , and by the theorem on a limit and its one-sided limits both one-sided limits at 1 are -2. If one of them were $\infty$ or $-\infty$ , it would not exist, by part (a) of Proposition 1; but it is -2. So neither one-sided limit is infinite, and the line $x = 1$ is not a vertical asymptote. The graph has a hole at $(1, -2)$ .
At 2: an asymptote. For $x$ in $\bigl(\frac32, \frac52\bigr)$ with $x \ne 2$ , step 2 gives $f(x) = (x + 1) \cdot \frac{1}{x - 2}$ , and adding 1 gives $x + 1 > \frac52$ (property 3). By parts (a) and (c) of Proposition 2 and part (a) of Proposition 3 with $m = \frac52$ , as in Example 1: $f(x) \to \infty$ as $x \to 2^{+}$ and $f(x) \to -\infty$ as $x \to 2^{-}$ .

\boxed{ \begin{aligned} &x = 2 \text{ is a vertical asymptote,} \\ &\lim_{x \to 1} f(x) = -2 \end{aligned} }

(15)

Check. $f(1.001) = \frac{0.002001}{-0.000999} \approx -2.003$ , close to ${-2}$ . ✓ $f(2.001) = \frac{3.004001}{0.001001} \approx 3001$ . ✓ For $\eps = 0.06$ : $\delta = 0.01$ , and $x = 1.009$ gives $\abs{f(x) + 2} = \frac{3 \cdot 0.009}{0.991} \approx 0.0272 < 0.06$ . ✓

An applied example¶

Example 4 (The image of an object near the focal point)

For the lens of Why this matters, $v(u) = \dfrac{10u}{u - 10}$ (in centimetres) for $u \ne 10$ . (a) Find $\lim_{u \to 10^{+}} v(u)$ and $\lim_{u \to 10^{-}} v(u)$ . (b) Find the largest $\delta > 0$ such that every object distance $u$ with $10 < u < 10 + \delta$ puts the image more than 10 metres behind the lens.

Split and bound. For $u \ne 10$ , $v(u) = 10u \cdot \frac{1}{u - 10}$ . If $9 < u < 11$ , multiplying by the positive number 10 keeps both inequalities (property 5(a) of the order rules): $10u > 90$ . So $h(u) = 10u$ satisfies $h(u) \ge 90$ on $(9, 10)$ and on $(10, 11)$ .
(a) By parts (a) and (c) of Proposition 2 ( $a = 10$ , $n = 1$ ), $\frac{1}{u - 10}$ tends to $\infty$ from the right and to $-\infty$ from the left. By part (a) of Proposition 3 and its left-hand version, $\lim_{u \to 10^{+}} v(u) = \infty$ and $\lim_{u \to 10^{-}} v(u) = -\infty$ , as the table suggested.
(b) Solve the inequality. Ten metres are 1000 cm. For $u > 10$ , the number $u - 10$ is positive, and so is its reciprocal $\frac{1}{u - 10}$ (property 6); so multiplying by either keeps an inequality (property 5(a)): $v(u) > 1000$ exactly when $10u > 1000(u - 10) = 1000u - 10\,000$ . Adding $10\,000 - 10u$ to both sides (property 3), this says $10\,000 > 990u$ , and multiplying by the positive number $\frac{1}{990}$ , $u < \frac{1000}{99}$ .
The largest $\delta$ . So $v(u) > 1000$ at every $u$ with $10 < u < \frac{1000}{99}$ , and $\delta = \frac{1000}{99} - 10 = \frac{10}{99}$ works. A larger $\delta$ does not: its half-window contains $u = \frac{1000}{99}$ , where $v(u) = \frac{10\,000/99}{10/99} = 1000$ , which is not more than 1000.

\boxed{ \begin{aligned} \lim_{u \to 10^{+}} v(u) &= \infty, \\ \lim_{u \to 10^{-}} v(u) &= -\infty, \\ \delta &= \frac{10}{99} \end{aligned} }

(16)

The object must be less than $\frac{10}{99} \approx 0.101$ cm, about a millimetre, beyond the focal point.

Check. $u = 10.1$ lies in the half-window ( $10.1 < 10.10101\ldots$ ), and $v(10.1) = \frac{101}{0.1} = 1010 > 1000$ . ✓ The table in Why this matters: $v(10.01) = 10\,010$ and $v(9.99) = -9990$ . ✓ Units: $v$ and $u$ in centimetres, $\delta$ in centimetres. ✓

Common mistakes¶

Arithmetic with

\infty

✗ Wrong: “ $\frac{1}{x} \to \infty$ and $\frac{1}{x^2} \to \infty$ as $x \to 0^{+}$ , so $\frac{1}{x} - \frac{1}{x^2} \to \infty - \infty = 0$ .”

Why: $\infty - \infty$ is not a number, and the two terms may grow at different rates. Here the second one wins.

✓ Right: combine first: $\frac{1}{x} - \frac{1}{x^2} = (x - 1) \cdot \frac{1}{x^2}$ for $x \ne 0$ . On $\bigl(0, \frac12\bigr)$ , $x - 1 < -\frac12$ , so part (b) of Proposition 3 and Proposition 2 give $\frac{1}{x} - \frac{1}{x^2} \to -\infty$ as $x \to 0^{+}$ (Exercise 9 has more).

Rigorous track¶

Why the factor must stay away from 0

Proposition 3 asks for $h(x) \ge m$ with a fixed $m > 0$ , not only $h(x) > 0$ . That is needed. Let $h(x) = x^2$ and $g(x) = \frac{1}{x}$ for $0 < x < 1$ . Then $h(x) > 0$ and $\lim_{x \to 0^{+}} g(x) = \infty$ , but $h(x)\,g(x) = x$ is less than 1 at every point of $(0, 1)$ , so no $\delta$ wins the round $M = 1$ for it, and it does not tend to $\infty$ . The values of $h$ come arbitrarily close to 0, and they beat the growth of $g$ . In the same way, part (c) needs $L \ne 0$ : with $h(x) = x$ , which tends to 0, the product $h(x)\,g(x) = 1$ is constant.

Unbounded is not the same as tending to

\infty

A function can take values beyond every height in every half-window, and still not tend to $\infty$ . For $x > 0$ , let $k(x)$ be the integer part of $\frac{1}{x}$ , and let $f(x) = \frac{1}{x}$ if $k(x)$ is even and $f(x) = 0$ if $k(x)$ is odd.

Let $\delta > 0$ . By the Archimedean property there is an integer $n \ge 1$ with $n > \frac{1}{\delta}$ , and then the even integer $2n$ and the odd integer $2n + 1$ are both greater than $\frac{1}{\delta}$ . For an integer $j > \frac{1}{\delta}$ , the point $x = \frac{1}{j}$ satisfies $0 < x < \delta$ , by property 6 of the order rules, and $\frac{1}{x} = j$ , whose integer part is $j$ itself (as $j \le j < j + 1$ ). So:

at $x = \frac{1}{2n + 1}$ , $f(x) = 0 < 1$ : every half-window contains a point that fails the round $M = 1$ , and $f$ does not tend to $\infty$ as $x \to 0^{+}$ ;
at $x = \frac{1}{2n}$ , $f(x) = 2n$ , and choosing $n$ also greater than a given $M$ makes $f(x) > M$ : the values of $f$ in every half-window are not bounded above.

Nor does $f$ tend to $-\infty$ : its values are never negative, so every point fails the round $M = 1$ . Nor does it have a real right-hand limit $L$ : with $\eps = 1$ , that would give a half-window on which $f(x) < L + 1$ (part (a) of the proposition on distance inequalities), but that half-window contains points $\frac{1}{2n}$ with $2n > L + 1$ , where $f$ is $2n$ . So $\lim_{x \to 0^{+}} f(x)$ fails to exist in a way that “ $= \infty$ ” does not describe.

The definition in symbols, and its negation

With quantifiers, $\lim_{x \to a^{+}} f(x) = \infty$ says

\begin{aligned} &\forall M > 0 \ \ \exists \delta > 0 \ \ \forall x \colon \\ &\quad a < x < a + \delta \implies \\ &\qquad x \in \dom f \\ &\qquad \text{and } f(x) > M . \end{aligned}

(17)

To negate, each $\forall$ becomes $\exists$ and each $\exists$ becomes $\forall$ , the implication $P \implies Q$ becomes “ $P$ and not $Q$ ”, and “not $(u > v)$ ” becomes “ $u \le v$ ” (property 7 of the order rules). So $f$ does not tend to $\infty$ as $x \to a^{+}$ when

\begin{aligned} &\exists M > 0 \ \ \forall \delta > 0 \ \ \exists x \colon \\ &\quad a < x < a + \delta \ \text{ and} \\ &\qquad \bigl(x \notin \dom f \\ &\qquad\quad \text{or } f(x) \le M\bigr). \end{aligned}

(18)

This is the pattern of the non-example after Definition 2 (in its left-hand version, with the half-windows $a - \delta < x < a$ ), with $M = 1$ , and of the function above, where the points $\frac{1}{2n + 1}$ fail the round $M = 1$ .

Summary¶

$\lim_{x \to a^{+}} f(x) = \infty$ (Definition 1): $f(x)$ is above every height for $x > a$ close enough to $a$ . Precisely (Definition 2): for every $M > 0$ there is a $\delta > 0$ such that
$\boxed{ \begin{aligned} &a < x < a + \delta \\ &\quad \implies f(x) > M ; \end{aligned} }$
(19)
for $-\infty$ , $f(x) < -M$ ; from the left, $a - \delta < x < a$ . The two-sided $\lim_{x \to a} f(x) = \infty$ means that both one-sided limits are $\infty$ .
“ $= \infty$ ” is notation: the limit does not exist as a real number (Proposition 1).
$\frac{1}{(x - a)^n} \to \infty$ as $x \to a^{+}$ ; as $x \to a^{-}$ it tends to $\infty$ for even $n$ and to $-\infty$ for odd $n$ (Proposition 2).
A factor that stays at least $m > 0$ away from 0, with one sign, keeps an infinite limit infinite and decides its sign (Proposition 3). For a rational function at a zero of the denominator: factor, read the signs, bound the other factor, conclude.
The line $x = a$ is a vertical asymptote if at least one one-sided limit at $a$ is $\infty$ or $-\infty$ (Definition 3). A zero of the denominator where the numerator is also 0 may be a hole instead.

Exercises¶

Exercise 6 (Two zeros of the denominator)

Let $f(x) = \dfrac{x - 2}{x^2 + x}$ . Find (a) $\displaystyle \lim_{x \to -1^{-}} f(x)$ , (b) $\displaystyle \lim_{x \to -1^{+}} f(x)$ , (c) $\displaystyle \lim_{x \to 0^{-}} f(x)$ and (d) $\displaystyle \lim_{x \to 0^{+}} f(x)$ .

Hint 1

$x^2 + x = x(x + 1)$ . Make a sign chart with the numbers ${-1}$ , 0 and 2.

Hint 2

Near ${-1}$ , write $f(x) = \frac{x - 2}{x} \cdot \frac{1}{x + 1}$ and bound the first factor on $\bigl(-\frac32, -\frac12\bigr)$ . Near 0, write $f(x) = \frac{x - 2}{x + 1} \cdot \frac{1}{x}$ and bound the first factor on $\bigl(-\frac12, \frac12\bigr)$ .

Answer

(a) $-\infty$ (b) $\infty$ (c) $\infty$ (d) $-\infty$

Solution to Exercise 6 #

$x^2 + x = x(x + 1)$ is 0 exactly at $x = 0$ and $x = -1$ (part (a) of the sign rules), and the numerator $x - 2$ is not 0 at either. We follow Example 2.

Sign. By Proposition 4, counting the negative factors among $x - 2$ , $x$ and $x + 1$ : on $(-\infty, -1)$ all three are negative, so $f < 0$ ; on $(-1, 0)$ two are, so $f > 0$ ; on $(0, 2)$ one is, so $f < 0$ . So we expect (a) $-\infty$ , (b) $\infty$ , (c) $\infty$ , (d) $-\infty$ .

Near ${-1}$ . For $x \ne 0, -1$ , $f(x) = h(x) \cdot \frac{1}{x + 1}$ with $h(x) = \frac{x - 2}{x} = \frac{2 - x}{-x}$ . Let $-\frac32 < x < -\frac12$ . Multiplying by ${-1}$ reverses both inequalities (property 5(c) of the order rules): $\frac12 < -x < \frac32$ , and adding 2 (property 3), $\frac52 < 2 - x < \frac72$ . By property 6, $\frac{1}{-x} > \frac23$ . Multiplying $2 - x > \frac52$ by the positive number $\frac{1}{-x}$ , and $\frac{1}{-x} > \frac23$ by the positive number $\frac52$ (property 5(a)), gives $h(x) > \frac53$ . With $\frac{1}{x + 1} \to \infty$ as $x \to -1^{+}$ and $-\infty$ as $x \to -1^{-}$ (Proposition 2, $n = 1$ ), part (a) of Proposition 3 and its left-hand version give (a) $-\infty$ and (b) $\infty$ .

Near 0. For $x \ne 0, -1$ , $f(x) = k(x) \cdot \frac{1}{x}$ with $k(x) = \frac{x - 2}{x + 1}$ . Let $-\frac12 < x < \frac12$ . Adding -2 and 1 (property 3): $x - 2 < -\frac32$ and $\frac12 < x + 1 < \frac32$ , so $\frac{1}{x + 1} > \frac23$ (property 6). Multiplying $x - 2 < -\frac32$ by the positive number $\frac{1}{x + 1}$ keeps it, and multiplying $\frac{1}{x + 1} > \frac23$ by the negative number $-\frac32$ reverses it (property 5(c)): $k(x) < -\frac32 \cdot \frac{1}{x + 1} < -\frac32 \cdot \frac23 = -1$ . With $\frac{1}{x} \to -\infty$ as $x \to 0^{-}$ and $\infty$ as $x \to 0^{+}$ , part (b) of Proposition 3, which flips the sign, gives (c) $\infty$ and (d) $-\infty$ .

All four agree with the sign chart. Both lines $x = -1$ and $x = 0$ are vertical asymptotes.

Solution to Exercise 7 #

Factor. $x^2 - 4 = (x - 2)(x + 2)$ and $x^2 - x - 2 = (x - 2)(x + 1)$ . The denominator is 0 exactly at $x = 2$ and $x = -1$ (part (a) of the sign rules). At every $x \ne 2, -1$ we may cancel the non-zero factor $x - 2$ : $f(x) = \frac{x + 2}{x + 1}$ .

At ${-1}$ : an asymptote. For $x \ne 2, -1$ , $f(x) = (x + 2) \cdot \frac{1}{x + 1}$ . If $-\frac32 < x < -\frac12$ , adding 2 gives $x + 2 > \frac12$ (property 3 of the order rules). By part (a) of Proposition 2 ( $a = -1$ , $n = 1$ ) and part (a) of Proposition 3 with $m = \frac12$ on $\bigl(-1, -\frac12\bigr)$ , $\lim_{x \to -1^{+}} f(x) = \infty$ . So $x = -1$ is a vertical asymptote.

At 2: a finite limit. $f$ is defined at every point of $(1, 3)$ except 2. For $x \ne 2, -1$ ,

\begin{aligned} f(x) - \frac43 &= \frac{3(x + 2) - 4(x + 1)}{3(x + 1)} \\ &= \frac{2 - x}{3(x + 1)}, \end{aligned}

(20)

so $\abs{f(x) - \frac43} = \frac{\abs{x - 2}}{3\abs{x + 1}}$ , by properties 2 and 4 of the absolute value. If $\abs{x - 2} < 1$ , then $1 < x < 3$ (part (a) of the proposition on distance inequalities), so $x + 1 > 2$ (adding 1, property 3 of the order rules) and $\abs{x + 1} = x + 1$ (the definition of the absolute value). Multiplying $x + 1 > 2$ by the positive number 3 keeps it (property 5(a)), so $3\abs{x + 1} > 6$ , and property 6 gives $\frac{1}{3\abs{x + 1}} < \frac16$ . Let $\eps > 0$ , $\delta = \min(1, 6\eps)$ and $0 < \abs{x - 2} < \delta$ . Then $f(x)$ is defined, and multiplying $\frac{1}{3\abs{x + 1}} < \frac16$ by the positive number $\abs{x - 2}$ keeps it (property 5(a)):

\abs{f(x) - \tfrac43} < \frac{\abs{x - 2}}{6} < \frac{6\eps}{6} = \eps .

(21)

So $\lim_{x \to 2} f(x) = \frac43$ . By the theorem on a limit and its one-sided limits, both one-sided limits at 2 are $\frac43$ , so by part (a) of Proposition 1 neither is $\infty$ or $-\infty$ . The line $x = 2$ is not a vertical asymptote; the graph has a hole at $\bigl(2, \frac43\bigr)$ .

So only the zero ${-1}$ of the denominator gives a vertical asymptote.

Exercise 8 (The cost of a cleaner river)

A town estimates that removing $p$ per cent of the pollution from its river costs $C(p) = \dfrac{50p}{100 - p}$ thousand pounds, for $0 \le p < 100$ . (a) Find $\displaystyle \lim_{p \to 100^{-}} C(p)$ . (b) Find the largest $\delta > 0$ such that every $p$ with $100 - \delta < p < 100$ has a cost of more than 4950 thousand pounds.

Hint 1

$\frac{1}{100 - p} = -\frac{1}{p - 100}$ . What does $\frac{1}{p - 100}$ do as $p \to 100^{-}$ , and what sign does $-50p$ have near 100?

Hint 2

For $p < 100$ , the number $100 - p$ is positive, so you may multiply the inequality $C(p) > 4950$ by it.

Answer

(a) $\infty$ (b) 1

Solution to Exercise 12 #

The denominator $(x - 1)(x - 3)$ is 0 exactly at 1 and 3 (part (a) of the sign rules). Let $g(x) = \frac{1}{(x - 1)(x - 3)}$ for $x \ne 1, 3$ ; then $f(x) = (x - c)\, g(x)$ at every such $x$ .

The one-sided limits of $g$ at 1. On $\bigl(\frac12, \frac32\bigr)$ , $g(x) = \frac{1}{x - 3} \cdot \frac{1}{x - 1}$ for $x \ne 1$ . There $\frac32 < 3 - x < \frac52$ (multiplying by ${-1}$ and adding 3, properties 5(c) and 3 of the order rules), so $\frac{1}{3 - x} > \frac25$ (property 6) and, multiplying by ${-1}$ , $\frac{1}{x - 3} < -\frac25$ . With Proposition 2 ( $n = 1$ ) and part (b) of Proposition 3 ( $m = \frac25$ ): $g(x) \to -\infty$ as $x \to 1^{+}$ and $g(x) \to \infty$ as $x \to 1^{-}$ .

The one-sided limits of $g$ at 3. On $\bigl(\frac52, \frac72\bigr)$ , $g(x) = \frac{1}{x - 1} \cdot \frac{1}{x - 3}$ for $x \ne 3$ , and $\frac32 < x - 1 < \frac52$ , so $\frac{1}{x - 1} > \frac25$ . By part (a) of Proposition 3: $g(x) \to \infty$ as $x \to 3^{+}$ and $g(x) \to -\infty$ as $x \to 3^{-}$ .

If $c \ne 1$ , then $x = 1$ is an asymptote. For every real $b$ , $\lim_{x \to b} (x - c) = b - c$ , since $\abs{(x - c) - (b - c)} = \abs{x - b}$ and $\delta = \eps$ works; by the theorem on a limit and its one-sided limits, both one-sided limits are $b - c$ too. With $b = 1$ , the limit $1 - c$ is not 0, so part (c) of Proposition 3 (on $\bigl(1, \frac32\bigr)$ , with $h(x) = x - c$ ) shows that $\lim_{x \to 1^{+}} f(x)$ is $\infty$ or $-\infty$ . In the same way, if $c \ne 3$ , then $x = 3$ is an asymptote. So for $c \ne 1, 3$ there are two.

If $c = 1$ . Then $x = 3$ is an asymptote, as $c \ne 3$ . At every $x \ne 1, 3$ we may cancel $x - 1$ : $f(x) = \frac{1}{x - 3}$ . We show $\lim_{x \to 1} f(x) = -\frac12$ . For $x \ne 1, 3$ , $f(x) + \frac12 = \frac{x - 1}{2(x - 3)}$ , so $\abs{f(x) + \frac12} = \frac{\abs{x - 1}}{2\abs{x - 3}}$ (property 4 of the absolute value). If $\abs{x - 1} < 1$ , then $0 < x < 2$ (part (a) of the proposition on distance inequalities). Multiplying $x < 2$ by ${-1}$ reverses it (property 5(c) of the order rules), and adding 3 keeps it (property 3), so $3 - x > 1$ . So $x - 3 < 0$ and $\abs{x - 3} = 3 - x$ (the definition of the absolute value); multiplying $3 - x > 1$ by the positive number 2 keeps it (property 5(a)), and property 6 gives $\frac{1}{2\abs{x - 3}} < \frac12$ . For $\eps > 0$ , $\delta = \min(1, 2\eps)$ and $0 < \abs{x - 1} < \delta$ , $f(x)$ is defined and $\abs{f(x) + \frac12} < \frac{\abs{x - 1}}{2} < \eps$ . So both one-sided limits at 1 are the real number $-\frac12$ (by the theorem), neither is infinite (part (a) of Proposition 1), and $x = 1$ is not an asymptote: exactly one asymptote.

If $c = 3$ . In the same way, $x = 1$ is an asymptote, and for $x \ne 1, 3$ , $f(x) = \frac{1}{x - 1}$ and $f(x) - \frac12 = \frac{3 - x}{2(x - 1)}$ , so $\abs{f(x) - \frac12} = \frac{\abs{x - 3}}{2\abs{x - 1}}$ (properties 4 and 2 of the absolute value, as $\abs{3 - x} = \abs{-(x - 3)} = \abs{x - 3}$ ). If $\abs{x - 3} < 1$ , then $2 < x < 4$ and $\abs{x - 1} = x - 1 > 1$ , so $\delta = \min(1, 2\eps)$ works again: $\lim_{x \to 3} f(x) = \frac12$ , and $x = 3$ is not an asymptote. Exactly one asymptote.

So exactly one of the two zeros gives a vertical asymptote when $c = 1$ or $c = 3$ .

Where this leads¶

These topics build on this one:

Limits at Infinity and Horizontal Asymptotes

Found an error on this page? Report it (the form opens with this page filled in).

Infinite limits at a point describe the vertical lines of a graph. The next page of the chapter lets $x$ itself grow without bound, which gives the horizontal ones.

Interval	point	negative factors	$f$
$(-\infty, 0)$	${-1}$	4	$+$
$(0, 1)$	$\frac12$	2	$+$
$(1, 3)$	2	1	$-$
$(3, \infty)$	4	0	$+$