Limit Laws - Maths: an open, verified university mathematics reference

Why this matters¶

On the page The Limit of a Function we proved that $\lim_{x \to 2} x^2 = 4$ from the definition, and even that took some care: we had to cap the window and bound a factor. Now consider

\lim_{x \to 2} \frac{x^3 - 1}{x^2 + 3} .

(1)

A table suggests an answer, but a direct proof with $\eps$ and $\delta$ would have to control the numerator and the denominator at the same time, with a new choice of $\delta$ for this one function. Doing that for every function we meet would be hopeless.

Figure 1:Graph of $y = \frac{x^3 - 1}{x^2 + 3}$ for $0 \le x \le 4$ , rising from about ${-0.33}$ at $x = 0$ to about 3.3 at $x = 4$ , with a point on the graph that a slider for $x$ moves. A table lists the values, rounded to six decimal places, at $x = 1.9, 1.99, 1.999$ : 0.886384, 0.988578, 0.998857, and at $x = 2.001, 2.01, 2.1$ : 1.001143, 1.011435, 1.114845. They approach 1 from both sides.

Try this: move the point towards $x = 2$ from both sides and read $f(x)$ . Then compute the numerator and the denominator at $x = 2$ separately. What do you notice?

The table suggests that the limit is $1 = \frac{7}{7}$ , the numerator’s value at 2 divided by the denominator’s. That is no accident. Instead of proving each limit from scratch, we prove a few rules once, from the definition, that say how limits combine: the limit of a sum is the sum of the limits, the limit of a quotient is the quotient of the limits, and so on. With these limit laws the limit above takes a few lines (Example 2), and for every polynomial the limit is its value at the point (Corollary 1).

Each law has hypotheses, and they matter. The stone’s average speed on The Limit of a Function, $\frac{5(1+h)^2 - 5}{h}$ , is a quotient whose denominator approaches 0: the quotient law says nothing about it. The last part of the main results is about such cases.

Combining functions¶

Let $f$ and $g$ be functions. Their sum $f + g$ , difference $f - g$ and product $fg$ are the functions with

$(f + g)(x) = f(x) + g(x)$ ,
$(f - g)(x) = f(x) - g(x)$ ,
$(fg)(x) = f(x)\,g(x)$ ,

defined at every $x$ at which both $f(x)$ and $g(x)$ are defined. Their quotient $f/g$ , with $(f/g)(x) = \frac{f(x)}{g(x)}$ , is defined at every such $x$ at which, in addition, $g(x) \ne 0$ . For a real number $c$ , the function $cf$ has $(cf)(x) = c\,f(x)$ , and $\sqrt{f}$ has the value $\sqrt{f(x)}$ at every $x$ at which $f(x)$ is defined and $f(x) \ge 0$ , by the square-root remark.

Why the laws should hold. If $f(x)$ is close to $L$ and $g(x)$ is close to $M$ , then $f(x) + g(x)$ should be close to $L + M$ . How close? The two errors can add up: if each of $f(x)$ and $g(x)$ is within 0.01 of its limit, then $f(x) + g(x)$ is within 0.02 of $L + M$ , by the triangle inequality. So to get $f(x) + g(x)$ within a tolerance $\eps$ of $L + M$ , we demand $\frac{\eps}{2}$ from each of $f$ and $g$ . The definition of the limit lets us demand any tolerance, $\frac{\eps}{2}$ included. That is the whole idea of the proof of the sum law below, and the other proofs refine it.

Example. From the definition, $\lim_{x \to 3} (2x - 1) = 5$ (the window $\delta = \frac{\eps}{2}$ works, because $\abs{(2x - 1) - 5} = 2\abs{x - 3}$ ) and $\lim_{x \to 3} 4x = 12$ ( $\delta = \frac{\eps}{4}$ works, because $\abs{4x - 12} = 4\abs{x - 3}$ ). The sum law gives $\lim_{x \to 3} (6x - 1) = 5 + 12 = 17$ without a new proof; Example 1 looks at the $\delta$ it produces.

Non-example. The laws need the limits of the parts to exist. For $f(x) = x$ and $g(x) = \frac{1}{x}$ , $x \ne 0$ , the product $f(x)\,g(x) = 1$ has the limit 1 at 0, but the product law cannot be used to find it, because $\frac{1}{x}$ has no limit at 0 (we show this in When the laws do not apply).

Main results¶

The laws are stated for two functions $f$ and $g$ with limits $L$ and $M$ at the same point $a$ . Each of them comes with the conclusion that the new function is defined near $a$ , so that the definition of a limit applies to it.

Theorem 1 (Limit laws)

Let $a$ be a real number.

(a) Constants and the identity. For every real number $c$ , $\lim_{x \to a} c = c$ ; and $\lim_{x \to a} x = a$ .

Now let $f$ and $g$ be functions, each defined at every point of an open interval containing $a$ , except possibly at $a$ itself, such that $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ , where $L$ and $M$ are real numbers. Then:

(b) Sum and difference. $\lim_{x \to a} \bigl(f(x) + g(x)\bigr) = L + M$ and $\lim_{x \to a} \bigl(f(x) - g(x)\bigr) = L - M$ .
(c) Product. $\lim_{x \to a} f(x)\,g(x) = LM$ . In particular, $\lim_{x \to a} c\,f(x) = cL$ for every real number $c$ .
(d) Quotient. If $M \ne 0$ , then $\lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{L}{M}$ .
(e) Power. For every integer $n \ge 1$ , $\lim_{x \to a} \bigl(f(x)\bigr)^n = L^n$ .
(f) Square root. If $L > 0$ , then $\lim_{x \to a} \sqrt{f(x)} = \sqrt{L}$ . If $L = 0$ , and there is an $r > 0$ such that $f(x) \ge 0$ for every $x$ with $0 < \abs{x - a} < r$ , then $\lim_{x \to a} \sqrt{f(x)} = 0$ .

In each of (b)–(f), the function whose limit is taken is defined at every point of some open interval containing $a$ , except possibly at $a$ , so that the definition of a limit applies to it.

Proof

We prove (a) and (b) here; (c) and (d) are proved in the rigorous track below, and (e) and (f) are sketched in the remark after this proof. For the sum the idea is to share the tolerance: we ask $f(x)$ to be within $\frac{\eps}{2}$ of $L$ and $g(x)$ within $\frac{\eps}{2}$ of $M$ , and the triangle inequality adds the two errors.

(a) The constant function $c$ is defined for every $x$ . Let $\eps > 0$ and take $\delta = 1$ . For every $x$ with $0 < \abs{x - a} < 1$ , $\abs{c - c} = 0 < \eps$ . So $\lim_{x \to a} c = c$ . The function $x$ is also defined for every $x$ . Let $\eps > 0$ and take $\delta = \eps$ . If $0 < \abs{x - a} < \delta$ , then $\abs{x - a} < \eps$ . So $\lim_{x \to a} x = a$ .

(b), the sum. Let $\eps > 0$ . Then $\frac{\eps}{2} > 0$ , so we may use it as the tolerance in the definition of a limit. Since $\lim_{x \to a} f(x) = L$ , there is a $\delta_1 > 0$ such that $f$ is defined and $\abs{f(x) - L} < \frac{\eps}{2}$ at every $x$ with $0 < \abs{x - a} < \delta_1$ . Since $\lim_{x \to a} g(x) = M$ , there is a $\delta_2 > 0$ such that $g$ is defined and $\abs{g(x) - M} < \frac{\eps}{2}$ at every $x$ with $0 < \abs{x - a} < \delta_2$ .

Let $\delta = \min(\delta_1, \delta_2)$ , which is positive, and let $0 < \abs{x - a} < \delta$ . Since $\delta \le \delta_1$ and $\delta \le \delta_2$ , also $\abs{x - a} < \delta_1$ and $\abs{x - a} < \delta_2$ , so $x$ lies in both windows. Hence $f(x)$ and $g(x)$ are defined, and so is $f(x) + g(x)$ . By part (a) of the triangle inequality, applied to the numbers $f(x) - L$ and $g(x) - M$ ,

\begin{aligned} \abs{\bigl(f(x) + g(x)\bigr) - (L + M)} &= \abs{\bigl(f(x) - L\bigr) + \bigl(g(x) - M\bigr)} \\ &\le \abs{f(x) - L} + \abs{g(x) - M} \\ &< \frac{\eps}{2} + \frac{\eps}{2} = \eps . \end{aligned}

(2)

The last step adds two strict inequalities, which gives a strict inequality.

So $f + g$ is defined at every $x$ with $0 < \abs{x - a} < \delta$ , which by part (e) of the proposition on distance inequalities is the open interval $(a - \delta, a + \delta)$ with $a$ removed; the definition applies to $f + g$ , and the display shows that $\delta$ wins the round $\eps$ . Since $\eps > 0$ was arbitrary, $\lim_{x \to a} \bigl(f(x) + g(x)\bigr) = L + M$ .

(b), the difference. The same $\delta$ works. The only change is in the first line of the display: $\bigl(f(x) - g(x)\bigr) - (L - M) = \bigl(f(x) - L\bigr) + \bigl(-(g(x) - M)\bigr)$ , and $\abs{-(g(x) - M)} = \abs{g(x) - M}$ by property 2 of the absolute value, so the same two bounds apply.

The power law is the product law used again and again, and the square-root law needs one identity and a look at the domain. Both are sketched here; what a sketch leaves out is said in it.

Remark 1 (Sketches for the power and square-root laws)

(e) Power. For $n = 1$ the law is the hypothesis $\lim_{x \to a} f(x) = L$ . For $n = 2$ , $\bigl(f(x)\bigr)^2 = f(x)\,f(x)$ , so part (c) with $g = f$ gives the limit $L \cdot L = L^2$ . For $n = 3$ , $\bigl(f(x)\bigr)^3 = \bigl(f(x)\bigr)^2 f(x)$ , so part (c) again gives $L^2 \cdot L = L^3$ , and so on: each step multiplies by one more factor $f(x)$ . What this leaves out: “and so on” stands for a proof by induction on $n$ , which shows that if the law holds for $n$ , it holds for $n + 1$ .

(f) Square root, $L > 0$ . At every $x$ with $f(x) \ge 0$ , the number $\sqrt{f(x)}$ is defined, and

\bigl(\sqrt{f(x)} - \sqrt{L}\bigr)\bigl(\sqrt{f(x)} + \sqrt{L}\bigr) = f(x) - L,

(3)

because $\bigl(\sqrt{y}\bigr)^2 = y$ for every $y \ge 0$ (the square-root remark). The second factor is at least $\sqrt{L}$ (add $\sqrt{L}$ to both sides of $\sqrt{f(x)} \ge 0$ ), and $\sqrt{L}$ is positive: $\sqrt{L} \ge 0$ , and $\sqrt{L} \ne 0$ because $0^2 = 0 \ne L$ . Dividing by the positive second factor, and using property 4 of the absolute value and, since the second factor is positive, the definition to drop the bars around it,

\abs{\sqrt{f(x)} - \sqrt{L}} = \frac{\abs{f(x) - L}}{\sqrt{f(x)} + \sqrt{L}} \le \frac{\abs{f(x) - L}}{\sqrt{L}} .

(4)

The inequality uses the order rules. The number $\sqrt{L}\,(\sqrt{f(x)} + \sqrt{L})$ is positive, as the product of two positive numbers (multiplying $0 < \sqrt{f(x)} + \sqrt{L}$ by $\sqrt{L} > 0$ , property 5(a)), so its reciprocal $\frac{1}{\sqrt{L}\,(\sqrt{f(x)} + \sqrt{L})}$ is positive (property 6). Multiplying $\sqrt{L} \le \sqrt{f(x)} + \sqrt{L}$ by this positive number keeps it (property 5(b)), and gives $\frac{1}{\sqrt{f(x)} + \sqrt{L}} \le \frac{1}{\sqrt{L}}$ . Multiplying this by $\abs{f(x) - L}$ , which is $\ge 0$ and may be 0, gives only $\le$ (property 5(b)). So a window in which $\abs{f(x) - L} < \eps\sqrt{L}$ also has $\abs{\sqrt{f(x)} - \sqrt{L}} < \eps$ : multiplying $\abs{f(x) - L} < \eps\sqrt{L}$ by the positive number $\frac{1}{\sqrt{L}}$ (property 6) keeps the strict inequality (property 5(a)), and $\le$ followed by $<$ gives $<$ (transitivity, property 2). What this leaves out: that $\sqrt{f(x)}$ is defined in the window at all, that is, $f(x) \ge 0$ there. The tolerance $L$ in the definition of $\lim_{x \to a} f(x) = L$ gives a window in which $\abs{f(x) - L} < L$ , so $f(x) > 0$ by part (a) of the proposition on distance inequalities; a full proof takes the smaller of the two windows.

(f) Square root, $L = 0$ . Here the hypothesis $f(x) \ge 0$ provides the domain, and by the order part of the square-root remark, $0 \le f(x) < \eps^2$ gives $\sqrt{f(x)} < \eps$ . So a window in which $\abs{f(x) - 0} < \eps^2$ and $f(x) \ge 0$ works. What this leaves out: the bookkeeping of the windows, as above.

Direct substitution¶

A polynomial is a function of the form $p(x) = c_n x^n + \dots + c_1 x + c_0$ , where $n \ge 0$ is an integer and $c_0, c_1, \dots, c_n$ are real numbers; it is defined for every real $x$ . A rational function is a quotient $\frac{p(x)}{q(x)}$ of two polynomials, defined at every $x$ with $q(x) \ne 0$ . For these functions the laws reduce every limit to a substitution.

Proof

We build $p$ from constants and the function $x$ with the laws of Theorem 1, one operation at a time.

(a) Write $p(x) = c_n x^n + \dots + c_1 x + c_0$ . Every term is defined for every real $x$ , so each satisfies the hypothesis on the domain in Theorem 1. By part (a), $\lim_{x \to a} x = a$ . For each $k$ with $1 \le k \le n$ , part (e) applied to the function $x$ gives $\lim_{x \to a} x^k = a^k$ , and then the “in particular” of part (c) gives $\lim_{x \to a} c_k x^k = c_k a^k$ . By part (a), $\lim_{x \to a} c_0 = c_0$ . The polynomial is the sum of these $n + 1$ terms; applying the sum law, part (b), $n$ times, adding one term at a time, gives

\lim_{x \to a} p(x) = c_n a^n + \dots + c_1 a + c_0 = p(a) .

(5)

(b) By (a), $\lim_{x \to a} p(x) = p(a)$ and $\lim_{x \to a} q(x) = q(a)$ , and both polynomials are defined for every real $x$ . Since $q(a) \ne 0$ , part (d) of Theorem 1 applies, with $f = p$ , $g = q$ , $L = p(a)$ and $M = q(a)$ , and gives $\lim_{x \to a} \frac{p(x)}{q(x)} = \frac{p(a)}{q(a)}$ . Since $r = \frac{p}{q}$ , this is $\lim_{x \to a} r(x)$ ; and since $q(a) \ne 0$ , $r$ is defined at $a$ , with $r(a) = \frac{p(a)}{q(a)}$ .

(c) Since $a$ is in $I$ and $r$ is defined at $a$ , so is $s$ , and $s(a) = r(a)$ . We find a window around $a$ that lies inside $I$ . By the definition of an interval, $I$ is $(\alpha, \beta)$ , $(\alpha, \infty)$ , $(-\infty, \beta)$ or $\R$ . Since $a$ is in $I$ , $\alpha < a$ when $I$ has the left endpoint $\alpha$ , and $a < \beta$ when it has the right endpoint $\beta$ . Let $\rho$ be the smaller of $a - \alpha$ and $\beta - a$ , or $a - \alpha$ , or $\beta - a$ , or 1, in these four cases. Then $\rho > 0$ : adding $-\alpha$ to both sides of $\alpha < a$ gives $0 < a - \alpha$ , and adding $-a$ to both sides of $a < \beta$ gives $0 < \beta - a$ (property 3 of the order rules), and the smaller of two positive numbers is one of them, so it is positive. Also $\rho \le a - \alpha$ when $I$ has the left endpoint $\alpha$ , and $\rho \le \beta - a$ when it has the right endpoint $\beta$ .

Let $\eps > 0$ . By (b), there is a $\delta_0 > 0$ such that $r$ is defined and $\abs{r(x) - r(a)} < \eps$ at every $x$ with $0 < \abs{x - a} < \delta_0$ . Let $\delta = \min(\delta_0, \rho)$ , which is positive, and let $0 < \abs{x - a} < \delta$ . By part (a) of the proposition on distance inequalities, $a - \delta < x < a + \delta$ . If $I$ has the left endpoint $\alpha$ , then adding $\alpha - \rho$ to both sides of $\rho \le a - \alpha$ gives $\alpha \le a - \rho$ , and adding $a - \rho - \delta$ to both sides of $\delta \le \rho$ gives $a - \rho \le a - \delta$ (both by property 3 of the order rules). So $\alpha \le a - \rho \le a - \delta < x$ , and $\alpha < x$ by transitivity (property 2). In the same way, if $I$ has the right endpoint $\beta$ , then $x < \beta$ . So $x$ is in $I$ . Since also $\abs{x - a} < \delta_0$ , $r$ is defined at $x$ , so $s$ is defined there, and

\abs{s(x) - s(a)} = \abs{r(x) - r(a)} < \eps .

(6)

So $s$ is defined at every $x$ with $0 < \abs{x - a} < \delta$ , which by part (e) of the proposition on distance inequalities is the open interval $(a - \delta, a + \delta)$ with $a$ removed; the definition of a limit applies to $s$ , and $\delta$ wins the round $\eps$ . Since $\eps > 0$ was arbitrary, $\lim_{x \to a} s(x) = s(a)$ .

In words. For a polynomial, the limit at any point is the value there. For a rational function, the same holds at every point where the denominator is not 0, and part (c) says that it still holds when the function is used only on an open interval around the point, as in an application where only some inputs make sense (Example 4). This is a property of these particular functions, proved from the laws. It is not true of every function given by a formula: the function $h$ of the remark on the value at $a$ , with $h(x) = x + 1$ for $x \ne 1$ and $h(1) = 5$ , has the limit 2 at 1, not $h(1) = 5$ .

Together with (e) and (f), the corollary handles many functions built from polynomials and square roots: apply it to the parts, check each law’s hypotheses, and combine (Example 3).

When the laws do not apply¶

Every law has hypotheses: the limits of the parts must exist, as real numbers; the quotient law needs $M \ne 0$ ; the square-root law needs $L > 0$ , or $L = 0$ and $f(x) \ge 0$ near $a$ . When a hypothesis fails, the law says nothing at all: neither that the limit exists, nor that it does not. Each of the following shows a hypothesis failing.

A part has no limit. The function $\frac{1}{x}$ , $x \ne 0$ , has no limit at 0. Suppose it had a limit $M$ . By part (a) of Theorem 1, $\lim_{x \to 0} x = 0$ , so the product law would give $\lim_{x \to 0} x \cdot \frac{1}{x} = 0 \cdot M = 0$ . But $x \cdot \frac{1}{x} = 1$ at every $x \ne 0$ , and the constant 1 is within every $\eps > 0$ of 1, so this limit is 1. Two different limits contradict uniqueness of limits. So $\frac{1}{x}$ has no limit at 0. In particular the product law does not apply to $x \cdot \frac{1}{x}$ , although its limit exists.
The quotient law with $M = 0$ and $L \ne 0$ . For $\frac{1}{x}$ at 0, the numerator has the limit 1 and the denominator the limit 0; the quotient has no limit, as we just saw. The same holds for $\frac{x + 1}{x}$ , which is $1 + \frac{1}{x}$ for $x \ne 0$ : if it had a limit, the difference law would give one for $\frac{1}{x} = \frac{x + 1}{x} - 1$ .
The quotient law with $M = 0$ and $L = 0$ . Substituting gives “ $\frac{0}{0}$ ”, which is not a number. Such a limit may be any number, or not exist. In the next figure, the numerator $x^2 + cx$ and the denominator $x$ both have the limit 0 at 0 (Corollary 1), for every value of $c$ .
Negative powers when $L = 0$ . Part (e) is stated for $n \ge 1$ . For $n = -1$ , $\bigl(f(x)\bigr)^{-1} = \frac{1}{f(x)}$ is a quotient with numerator 1, and when $L \ne 0$ the quotient law gives it the limit $\frac{1}{L}$ . What fails when $L = 0$ is the quotient law’s hypothesis $M \ne 0$ , here with $M = L$ , not anything about $n$ : for $f(x) = x$ at 0 there is no limit.
The square-root law without its sign condition. If $L < 0$ : $f(x) = x - 1$ has the limit ${-1}$ at 0, and $\sqrt{x - 1}$ is undefined at every $x < 1$ , so it is not defined on any open interval around 0, and the definition of a limit does not apply. If $L = 0$ but $f$ takes negative values arbitrarily close to $a$ : $f(x) = x$ at 0, where $\sqrt{x}$ is undefined at every $x < 0$ , and again the definition does not apply. With the sign condition, $f(x) = x^2 \ge 0$ , the law gives $\lim_{x \to 0} \sqrt{x^2} = 0$ .

Figure 2:Graph of $y = \frac{x^2 + cx}{x}$ for $-1 \le x \le 1$ , $x \ne 0$ , with a hole at $x = 0$ , where the formula is undefined. A slider sets $c$ from ${-2}$ to 2 in steps of 0.5, starting at $c = 1$ . For every $c$ the graph is a straight line with a hole. A table lists the values at $x = 0.1, 0.01, 0.001, -0.001, -0.01, -0.1$ ; for $c = 1$ they are 1.1, 1.01, 1.001, 0.999, 0.99, 0.9, which approach 1, and for $c = 1.5$ they are 1.6, 1.51, 1.501, 1.499, 1.49, 1.4, which approach 1.5.

Try this: set $c = -2$ , $c = 0$ and $c = 1.5$ . In each case, what do the numerator and the denominator approach as $x \to 0$ ? Which number do the values in the table approach? What does “ $\frac{0}{0}$ ” tell you about the limit?

The figure shows the same “ $\frac{0}{0}$ ” next to different values in the table, one for each $c$ . So a $\frac{0}{0}$ is not an answer: it says only that the quotient law does not apply, and that more work is needed.

Worked examples¶

Using the laws¶

Example 1 (The

\delta

that the proof of the sum law builds)

Let $f(x) = 2x - 1$ and $g(x) = 4x$ , with $\lim_{x \to 3} f(x) = 5$ and $\lim_{x \to 3} g(x) = 12$ . Follow the proof of the sum law, part (b) of Theorem 1, to find a $\delta$ for $\lim_{x \to 3} \bigl(f(x) + g(x)\bigr) = 17$ , and compare it with the largest $\delta$ that works.

Tolerance $\frac{\eps}{2}$ for $f$ . $\abs{f(x) - 5} = \abs{2x - 6} = 2\abs{x - 3}$ , so $\abs{f(x) - 5} < \frac{\eps}{2}$ exactly when $\abs{x - 3} < \frac{\eps}{4}$ (dividing by the positive number 2 keeps the inequality). Take $\delta_1 = \frac{\eps}{4}$ .
Tolerance $\frac{\eps}{2}$ for $g$ . $\abs{g(x) - 12} = \abs{4x - 12} = 4\abs{x - 3}$ , so $\abs{g(x) - 12} < \frac{\eps}{2}$ exactly when $\abs{x - 3} < \frac{\eps}{8}$ (dividing by the positive number 4). Take $\delta_2 = \frac{\eps}{8}$ .
The proof’s $\delta$ . $\delta = \min\bigl(\frac{\eps}{4}, \frac{\eps}{8}\bigr) = \frac{\eps}{8}$ , since $\frac{\eps}{8} < \frac{\eps}{4}$ for $\eps > 0$ .
Directly. $f(x) + g(x) = 6x - 1$ , and $\abs{(6x - 1) - 17} = 6\abs{x - 3}$ , which is less than $\eps$ exactly when $\abs{x - 3} < \frac{\eps}{6}$ (dividing by the positive number 6). So $\frac{\eps}{6}$ is the largest $\delta$ that works: a larger window contains $x = 3 + \frac{\eps}{6}$ , where the distance is exactly $\eps$ .

The proof gives

\boxed{\delta = \frac{\eps}{8}},

(7)

and the largest $\delta$ that works is $\frac{\eps}{6}$ . The proof’s $\delta$ is smaller than the largest one because the proof gives each function half of the tolerance, and $g$ needs more than half. On the largest window, $\abs{x - 3} < \frac{\eps}{6}$ , the error $\abs{g(x) - 12} = 4\abs{x - 3}$ comes as close as we like to $\frac{2\eps}{3}$ , while $\abs{f(x) - 5} = 2\abs{x - 3}$ stays below $\frac{\eps}{3}$ . Held to half of the tolerance, $g$ allows only the smaller window $\delta_2 = \frac{\eps}{8}$ . A proof only needs some $\delta$ that works.

Check. For $\eps = 0.6$ : $\delta = 0.075$ , and $x = 3.07$ gives $\abs{6(3.07) - 1 - 17} = 0.42 < 0.6$ . ✓

Example 2 (The limit from Why this matters)

Find $\displaystyle \lim_{x \to 2} \frac{x^3 - 1}{x^2 + 3}$ with the laws of Theorem 1, naming the law used at each step.

The denominator. By part (a), $\lim_{x \to 2} x = 2$ , so by part (e), $\lim_{x \to 2} x^2 = 4$ . By part (a), $\lim_{x \to 2} 3 = 3$ . By part (b), $\lim_{x \to 2} (x^2 + 3) = 4 + 3 = 7$ .
The hypothesis of the quotient law. The denominator’s limit is $7 \ne 0$ , so part (d) may be used.
The numerator. By part (e), $\lim_{x \to 2} x^3 = 2^3 = 8$ , and by parts (a) and (b), $\lim_{x \to 2} (x^3 - 1) = 8 - 1 = 7$ .
The quotient. By part (d), the limit is $\frac{7}{7} = 1$ .

\boxed{\lim_{x \to 2} \frac{x^3 - 1}{x^2 + 3} = 1}

(8)

These steps are the proof of Corollary 1 for this one function. From now on we may cite the corollary instead: the function is rational, its denominator is $7 \ne 0$ at $x = 2$ , and its value there is $\frac{7}{7} = 1$ .

Check. At $x = 2.001$ the function is $\frac{7.012006001}{7.004001} \approx 1.001143$ , as in the table of Why this matters. ✓

Example 3 (A square root inside a quotient)

Find $\displaystyle \lim_{x \to 1} \frac{x + \sqrt{x + 3}}{x^2 + 1}$ .

The square root. By Corollary 1, $\lim_{x \to 1} (x + 3) = 4$ . Since $4 > 0$ , part (f) of Theorem 1 gives $\lim_{x \to 1} \sqrt{x + 3} = \sqrt{4} = 2$ , and says that $\sqrt{x + 3}$ is defined near 1 (it is defined for every $x \ge -3$ ).
The numerator. By parts (a) and (b), $\lim_{x \to 1} \bigl(x + \sqrt{x + 3}\bigr) = 1 + 2 = 3$ .
The denominator. By Corollary 1, $\lim_{x \to 1} (x^2 + 1) = 2$ , which is not 0.
The quotient. By part (d), the limit is $\frac{3}{2}$ .

\boxed{\lim_{x \to 1} \frac{x + \sqrt{x + 3}}{x^2 + 1} = \frac{3}{2}}

(9)

The corollary alone does not give this limit: the function is not rational, because of the square root. The laws handle it one part at a time.

Check. At $x = 1.01$ : $\sqrt{4.01} \approx 2.002498$ , so the function is about $\frac{3.012498}{2.0201} \approx 1.491262$ , close to 1.5. ✓

Example 4 (Two resistors in parallel)

Two resistors in parallel, of 3 ohms and $t$ ohms, have the combined resistance

R(t) = \frac{3t}{3 + t}

(10)

ohms, for $t > 0$ . The second resistor warms up, and its resistance $t$ approaches 6 ohms. What does the combined resistance approach?

The function. Only $t > 0$ makes sense here, so $R$ is the restriction of the rational function $r(t) = \frac{3t}{3 + t}$ , defined at every $t \ne -3$ , to the open interval $(0, \infty)$ . That interval contains 6, and $R$ is defined at every point of it.
The hypothesis. The denominator at $t = 6$ is $3 + 6 = 9 \ne 0$ .
Substitute. By part (c) of Corollary 1, with $I = (0, \infty)$ , $\lim_{t \to 6} R(t) = R(6) = \frac{18}{9} = 2$ .

\boxed{\lim_{t \to 6} R(t) = 2}

(11)

So the combined resistance approaches 2 ohms.

Check. At $t = 6.01$ : $R(6.01) = \frac{18.03}{9.01} \approx 2.00111$ , in ohms. ✓ Units: ohms times ohms divided by ohms gives ohms. ✓ A parallel combination is smaller than each of its resistors, and $2 < 3$ . ✓

Checking the hypotheses¶

Example 5 (Do the laws apply?)

For each limit, decide whether the laws of Theorem 1 give it.

(i) $\displaystyle \lim_{x \to 1} \frac{x^2 - 1}{x - 1}$
(ii) $\displaystyle \lim_{x \to 0} x \cdot \frac{1}{x}$
(iii) $\displaystyle \lim_{x \to 2} \sqrt{x - 2}$

(i) By Corollary 1, the numerator and the denominator have the limits $1 - 1 = 0$ and $1 - 1 = 0$ . The quotient law needs the denominator’s limit to be non-zero, so it does not apply. The laws say nothing here: this is a “ $\frac{0}{0}$ ”, the subject of the next page.
(ii) The product law needs both factors to have limits, and $\frac{1}{x}$ has none at 0 (When the laws do not apply). So the law does not apply. The limit exists anyway: $x \cdot \frac{1}{x} = 1$ at every $x \ne 0$ , so every $\delta > 0$ wins every round, and the limit is 1.
(iii) The inside $x - 2$ has the limit 0, but it is negative at every $x < 2$ , so the sign condition of part (f) fails, and $\sqrt{x - 2}$ is undefined at every $x < 2$ . It is not defined on any open interval around 2, so the definition of a limit does not apply at all.

Conclusion. The laws give none of the three: (i) is a $\frac{0}{0}$ ; in (ii) a factor has no limit, yet the limit is 1; in (iii) the function is undefined to the left of 2.

“The laws do not apply” is not “the limit does not exist”: in (ii) the limit exists. When a law does not apply, we need another argument.

Check. (i) At $x = 1.001$ the numerator is 0.002001 and the denominator 0.001: both are close to 0. ✓ (ii) At $x = 0.001$ : $0.001 \cdot 1000 = 1$ . ✓ (iii) At $x = 1.99$ : $x - 2 = {-0.01}$ , which has no real square root. ✓

Common mistakes¶

Substituting into any formula

✗ Wrong: for the function $h$ of the remark on the value at $a$ , with $h(x) = x + 1$ for $x \ne 1$ and $h(1) = 5$ : “by direct substitution, $\lim_{x \to 1} h(x) = h(1) = 5$ .”

Why: Corollary 1 is about polynomials and rational functions. The function $h$ is neither. By the ✓ part below, its limit at 1 is 2. If $h$ were a polynomial, or a rational function (whose denominator is not 0 at 1, since $h(1)$ is defined), the corollary would give the limit $h(1) = 5$ , and two different limits contradict uniqueness of limits.

✓ Right: $h$ is the polynomial $x + 1$ with its value at 1 changed, and by the remark on the value at $a$ that does not change the limit. So $\lim_{x \to 1} h(x) = \lim_{x \to 1} (x + 1) = 2$ , by Corollary 1.

Rigorous track¶

The product and quotient laws need more than sharing the tolerance: in a product, the error in one factor is multiplied by the other factor, and in a quotient we divide by $g(x)$ , which must stay away from 0.

Proof (Rigorous track)

This proves parts (c) and (d) of Theorem 1. For the product we split $f(x)\,g(x) - LM$ into two terms, each a small error times a bounded factor; for the quotient we first show that $\frac{1}{g(x)}$ approaches $\frac{1}{M}$ and then use the product law. Every multiplication of an inequality below names the sign of the factor, by the order rules of Absolute Value and Inequalities (“Working with inequalities”).

(c) Product. The strategy rests on the identity

f(x)\,g(x) - LM = f(x)\bigl(g(x) - M\bigr) + M\bigl(f(x) - L\bigr),

(12)

which holds because the two middle terms, $-f(x)M$ and $+Mf(x)$ , cancel. Let $\eps > 0$ .

$f$ is bounded near $a$ . The definition of $\lim_{x \to a} f(x) = L$ , with the tolerance 1, gives a $\delta_1 > 0$ such that $f$ is defined and $\abs{f(x) - L} < 1$ at every $x$ with $0 < \abs{x - a} < \delta_1$ . For such $x$ , by part (a) of the triangle inequality, applied to the numbers $f(x) - L$ and $L$ , and then adding $\abs{L}$ to both sides of $\abs{f(x) - L} < 1$ ,
$\abs{f(x)} = \abs{\bigl(f(x) - L\bigr) + L} \le \abs{f(x) - L} + \abs{L} < 1 + \abs{L} .$
(13)
The tolerances. Let $K = \abs{L} + 1$ and $N = \abs{M} + 1$ . Both are at least 1, by property 1 of the absolute value, so $\frac{\eps}{2K}$ and $\frac{\eps}{2N}$ are positive. So there are $\delta_2 > 0$ and $\delta_3 > 0$ such that $g$ is defined and $\abs{g(x) - M} < \frac{\eps}{2K}$ at every $x$ with $0 < \abs{x - a} < \delta_2$ , and $\abs{f(x) - L} < \frac{\eps}{2N}$ at every $x$ with $0 < \abs{x - a} < \delta_3$ . (We use $N = \abs{M} + 1$ rather than $\abs{M}$ because $M$ may be 0, and we may not divide by 0.)
The proof. Let $\delta = \min(\delta_1, \delta_2, \delta_3)$ , which is positive, and let $0 < \abs{x - a} < \delta$ . Then $x$ lies in all three windows, so $f(x)$ , $g(x)$ and $f(x)\,g(x)$ are defined. By the identity, part (a) of the triangle inequality and property 4 of the absolute value,
$\begin{aligned} \abs{f(x)\,g(x) - LM} &= \abs{f(x)\bigl(g(x) - M\bigr) + M\bigl(f(x) - L\bigr)} \\ &\le \abs{f(x)}\,\abs{g(x) - M} + \abs{M}\,\abs{f(x) - L} . \end{aligned}$
(14)
The first term. By step 1, $\abs{f(x)} < K$ . Multiplying by $\abs{g(x) - M}$ , which is $\ge 0$ and may be 0, gives only $\abs{f(x)}\,\abs{g(x) - M} \le K\abs{g(x) - M}$ . Multiplying $\abs{g(x) - M} < \frac{\eps}{2K}$ by the positive number $K$ keeps the strict inequality: $K\abs{g(x) - M} < \frac{\eps}{2}$ . So the first term is less than $\frac{\eps}{2}$ .
The second term. $\abs{M} < N$ . Multiplying by $\abs{f(x) - L} \ge 0$ gives $\abs{M}\,\abs{f(x) - L} \le N\abs{f(x) - L}$ , and multiplying $\abs{f(x) - L} < \frac{\eps}{2N}$ by the positive number $N$ gives $N\abs{f(x) - L} < \frac{\eps}{2}$ . So the second term is less than $\frac{\eps}{2}$ .
Adding the two strict inequalities, $\abs{f(x)\,g(x) - LM} < \eps$ .

The product $fg$ is defined at every $x$ with $0 < \abs{x - a} < \delta$ , which by part (e) of the proposition on distance inequalities is the open interval $(a - \delta, a + \delta)$ with $a$ removed, so the definition applies to it, and $\lim_{x \to a} f(x)\,g(x) = LM$ . For the “in particular”, apply this to the constant function $c$ , which is defined everywhere and has the limit $c$ by part (a), and to $f$ : $\lim_{x \to a} c\,f(x) = cL$ .

(d) Quotient. Suppose $M \ne 0$ .

$g$ stays away from 0 near $a$ . By property 1 of the absolute value, $\abs{M} > 0$ , so $\frac{\abs{M}}{2}$ is a tolerance. The definition gives a $\delta_4 > 0$ such that $g$ is defined and $\abs{g(x) - M} < \frac{\abs{M}}{2}$ at every $x$ with $0 < \abs{x - a} < \delta_4$ . For such $x$ ,
$\abs{M} - \abs{g(x)} \le \abs{\abs{M} - \abs{g(x)}} \le \abs{M - g(x)} < \frac{\abs{M}}{2},$
(15)
by property 3 of the absolute value ( $u \le \abs{u}$ ), part (c) of the triangle inequality, and $\abs{M - g(x)} = \abs{g(x) - M}$ (property 2). Adding $\abs{g(x)} - \frac{\abs{M}}{2}$ to both sides gives $\abs{g(x)} > \frac{\abs{M}}{2} > 0$ . So $g(x) \ne 0$ , since $\abs{0} = 0$ by the definition. Hence $\frac{1}{g}$ is defined at every $x$ with $0 < \abs{x - a} < \delta_4$ , which by part (e) of the proposition on distance inequalities is the open interval $(a - \delta_4, a + \delta_4)$ with $a$ removed.
$\frac{1}{g(x)}$ approaches $\frac{1}{M}$ . Let $\eps > 0$ . The number $\frac{\eps\abs{M}^2}{2}$ is positive, so there is a $\delta_5 > 0$ such that $\abs{g(x) - M} < \frac{\eps\abs{M}^2}{2}$ at every $x$ with $0 < \abs{x - a} < \delta_5$ . Let $\delta = \min(\delta_4, \delta_5)$ and $0 < \abs{x - a} < \delta$ . Since $\frac{1}{g(x)} - \frac{1}{M} = \frac{M - g(x)}{g(x)\,M}$ , properties 4 and 2 of the absolute value give
$\abs{\frac{1}{g(x)} - \frac{1}{M}} = \frac{\abs{g(x) - M}}{\abs{g(x)}\,\abs{M}} .$
(16)
Multiplying $\abs{g(x)} > \frac{\abs{M}}{2}$ (step 1) by the positive number $\abs{M}$ gives $\abs{g(x)}\,\abs{M} > \frac{\abs{M}^2}{2} > 0$ . For positive numbers $u > v$ , multiplying by the positive number $\frac{1}{uv}$ gives $\frac{1}{v} > \frac{1}{u}$ ; so $\frac{1}{\abs{g(x)}\,\abs{M}} < \frac{2}{\abs{M}^2}$ . Multiplying this by $\abs{g(x) - M}$ , which is $\ge 0$ and may be 0, gives only $\le$ , and then multiplying $\abs{g(x) - M} < \frac{\eps\abs{M}^2}{2}$ by the positive number $\frac{2}{\abs{M}^2}$ keeps the strict inequality:
$\abs{\frac{1}{g(x)} - \frac{1}{M}} \le \frac{2\abs{g(x) - M}}{\abs{M}^2} < \eps .$
(17)
So $\lim_{x \to a} \frac{1}{g(x)} = \frac{1}{M}$ .
The quotient. By step 1, $\frac{1}{g}$ satisfies the hypothesis on the domain in the theorem, and so does $f$ . By step 2 and part (c), applied to $f$ and $\frac{1}{g}$ ,
$\lim_{x \to a} f(x) \cdot \frac{1}{g(x)} = L \cdot \frac{1}{M} = \frac{L}{M} .$
(18)
The functions $f \cdot \frac{1}{g}$ and $\frac{f}{g}$ are defined at the same $x$ (those where $f(x)$ and $g(x)$ are defined and $g(x) \ne 0$ ) and have the same values there, so they are the same function, and $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$ .

Sharing the tolerance in other ways

The proof of the sum law gives each function half of the tolerance, but any split works: if $\abs{f(x) - L} < \eps_1$ and $\abs{g(x) - M} < \eps_2$ with $\eps_1 + \eps_2 = \eps$ , the same display gives $\abs{(f(x) + g(x)) - (L + M)} < \eps$ . In Example 1, giving $\frac{2\eps}{3}$ to $f$ and $\frac{\eps}{3}$ to $g$ produces $\delta = \min\bigl(\frac{\eps}{3}, \frac{\eps}{12}\bigr) = \frac{\eps}{12}$ , a worse $\delta$ , and giving $\frac{\eps}{3}$ to $f$ and $\frac{2\eps}{3}$ to $g$ produces $\min\bigl(\frac{\eps}{6}, \frac{\eps}{6}\bigr) = \frac{\eps}{6}$ , the largest one.

For a sum of three functions, give each $\frac{\eps}{3}$ , or apply the law twice: $f + g + h = (f + g) + h$ . The proof of Corollary 1 does the second, once for each term of the polynomial.

Summary¶

The limit laws (Theorem 1) compute the limit of a sum, difference, product, quotient, power or square root from the limits of its parts, which must exist as real numbers.
The sum law is proved by sharing the tolerance: $\frac{\eps}{2}$ for each part, and $\delta = \min(\delta_1, \delta_2)$ .
The quotient law needs $M \ne 0$ ; the square-root law needs $L > 0$ , or $L = 0$ and $f(x) \ge 0$ near $a$ .
For a polynomial, and for a rational function at a point where its denominator is not 0, the limit is the value at the point (Corollary 1); this still holds when the rational function is restricted to an open interval around the point.
When a hypothesis fails, the law says nothing: “ $\frac{0}{0}$ ” is a signal to do more work, not an answer, and “the laws do not apply” is not “the limit does not exist”.

Exercises¶

Exercise 7 (The

\delta

from the proof of the sum law)

Let $f(x) = 3x - 2$ and $g(x) = 5 - x$ , so that $\lim_{x \to 2} f(x) = 4$ and $\lim_{x \to 2} g(x) = 3$ . Let $\eps > 0$ . (a) In the proof of the sum law, part (b) of Theorem 1, take for $\delta_1$ and $\delta_2$ the largest $\delta$ that works for $f$ and for $g$ with the tolerance $\frac{\eps}{2}$ . Which $\delta$ does the proof give for $f + g$ ? (b) What is the largest $\delta$ that works for $\lim_{x \to 2} \bigl(f(x) + g(x)\bigr) = 7$ with the tolerance $\eps$ ?

Hint 1

Write $\abs{f(x) - 4}$ , $\abs{g(x) - 3}$ and $\abs{(f(x) + g(x)) - 7}$ as multiples of $\abs{x - 2}$ , as in Example 1.

Answer

(a) $\frac{\eps}{6}$ (b) $\frac{\eps}{2}$

Where this leads¶

These topics build on this one:

Computing Limits Algebraically
The Squeeze Theorem
Limits at Infinity and Horizontal Asymptotes
Continuity (coming soon)

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From here on, limits are computed with the laws, and the definition of a limit is needed only to prove new laws. The next pages extend the toolkit: rewriting “ $\frac{0}{0}$ ” forms, the squeeze theorem, and limits at infinity.