Limits

(a) The denominator $q(x) = x^2 + 1$ has $q(1) = 2 \ne 0$, so by part (b) of direct substitution the limit is

A numerator that approaches 0 does no harm while the denominator does not.

(b) Here the denominator is $1 - 1 = 0$ at $x = 1$, so neither direct substitution nor the quotient law, part (d) of the limit laws, applies. The numerator is 0 at 1 too. Multiplying out shows that

Let $0 < \abs{x - 1} < 2$. By part (e) of the proposition on distance inequalities, $-1 < x < 1$ or $1 < x < 3$, so $x \ne 1$ and $x \ne -1$. Then $x - 1 \ne 0$ and $x + 1 \ne 0$, the quotient is defined, and cancelling the non-zero factor $x - 1$ gives

The right-hand side is a rational function whose denominator is $2 \ne 0$ at 1, so by part (b) of direct substitution its limit at 1 is $\frac{4}{2} = 2$. The two functions are defined and equal at every $x$ with $0 < \abs{x - 1} < 2$, so by the lemma on functions that agree except at a point, with $r = 2$, the limit in (b) is 2 as well.

The denominator is $x^2 + 3x = x(x + 3)$, which is 0 exactly when $x = 0$ or $x = -3$, by part (a) of the sign rules. So the function is defined at every $x \ne 0, -3$, and for every such $x$

The right-hand side is defined at exactly the same $x$, so the two sides are the same function. Both factors are defined at every $x$ with $0 < \abs{x} < 3$. By the theorem on the limit of $\frac{\sin x}{x}$, the first factor approaches 1. The second is a rational function whose denominator is $3 \ne 0$ at 0, so by part (b) of direct substitution it approaches $\frac{1}{3}$. By the product law, part (c) of the limit laws,

On each side. For every $x$, $x^2 - 4 = (x - 2)(x + 2)$. By part (b) of the sign rules, $x - 2 > 0$ for $x > 2$ and $x - 2 < 0$ for $x < 2$, so by the definition of the absolute value, $\abs{x - 2} = x - 2$ for $x > 2$ and $\abs{x - 2} = -(x - 2)$ for $x < 2$. Cancelling the non-zero number $x - 2$, $f(x) = x + 2$ for $x > 2$, and $f(x) = -(x + 2)$ for $x < 2$.

The one-sided limits of the formulas. The polynomials $x + 2$ and $-(x + 2)$ have the limits 4 and ${-4}$ at 2, by part (a) of direct substitution. They are defined at every real number, so by the theorem on a limit and its one-sided limits, $\lim_{x \to 2^{+}} (x + 2) = 4$ and $\lim_{x \to 2^{-}} \bigl(-(x + 2)\bigr) = -4$.

From the formulas to $f$. Let $\eps > 0$. By part (a) of the definition of one-sided limits, some $\delta > 0$ has $\abs{(x + 2) - 4} < \eps$ whenever $2 < x < 2 + \delta$. Every such $x$ has $x > 2$, so $f(x)$ is defined and equals $x + 2$, and $\abs{f(x) - 4} < \eps$. So the same $\delta$ wins the round $\eps$ for $f$, and $\lim_{x \to 2^{+}} f(x) = 4$. In the same way, with part (b) of that definition and the half-window $2 - \delta < x < 2$, where $f(x) = -(x + 2)$, $\lim_{x \to 2^{-}} f(x) = -4$.

The limit. $f$ is defined at every point of the open interval $(1, 3)$ except 2, so it is as in the definition of the limit. Its one-sided limits at 2 are ${-4}$ and 4, which differ, so $\lim_{x \to 2} f(x)$ does not exist, by part (a) of the corollary on one-sided limits that differ. By the remark on the jump, $f$ jumps by $4 - (-4) = 8$ at 2.

Factor. $2x^2 + 2x = 2x(x + 1)$ and $x^2 - 2x - 3 = (x + 1)(x - 3)$. The denominator is 0 exactly at $x = -1$ and $x = 3$, by part (a) of the sign rules. At every $x \ne -1, 3$ we may cancel the non-zero factor $x + 1$:

At 3: an asymptote. Let $3 < x < 4$. Then $x \ne -1$, so $f(x) = 2x \cdot \frac{1}{x - 3}$. Multiplying $x > 3$ by the positive number 2 keeps it (property 5(a) of the order rules), so $2x > 6$. By part (a) of the proposition on reciprocal powers, with $a = 3$ and $n = 1$, $\lim_{x \to 3^{+}} \frac{1}{x - 3} = \infty$. By part (a) of the proposition on a factor that stays away from 0, with $h(x) = 2x \ge 6 = m$ on $(3, 4)$, $\lim_{x \to 3^{+}} f(x) = \infty$. So the line $x = 3$ is a vertical asymptote, by the definition of a vertical asymptote.

At ${-1}$: a hole. Let $0 < \abs{x + 1} < 1$. By part (e) of the proposition on distance inequalities, $-2 < x < -1$ or $-1 < x < 0$, so $x \ne 3$, $f(x)$ is defined, and $f(x) = \frac{2x}{x - 3}$. This rational function has the denominator ${-4} \ne 0$ at ${-1}$, so by part (b) of direct substitution its limit at ${-1}$ is $\frac{-2}{-4} = \frac{1}{2}$. By the lemma on functions that agree except at a point, with $r = 1$, $\lim_{x \to -1} f(x) = \frac12$. Then, by the theorem on a limit and its one-sided limits, both one-sided limits of $f$ at ${-1}$ are the real number $\frac12$. By part (a) of the proposition that an infinite limit is not a real limit, a one-sided limit that is $\infty$ or $-\infty$ is no real number, so neither one-sided limit is $\infty$ or $-\infty$, and the line $x = -1$ is not a vertical asymptote. The graph has a hole at $\bigl({-1}, \frac12\bigr)$.

Horizontal asymptotes. $f = \frac{p}{q}$, where $p(x) = 2x^2 + 2x$ has degree 2 and leading coefficient 2, and $q(x) = x^2 - 2x - 3$ has degree 2 and leading coefficient 1. By part (c) of the proposition on rational functions at infinity, $\lim_{x \to \infty} f(x) = \frac21 = 2$ and $\lim_{x \to -\infty} f(x) = 2$. So $y = 2$ is a horizontal asymptote, by the definition of a horizontal asymptote. If $y = L$ is one, then $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$, and in either case $L = 2$, by uniqueness, property 1 of the remark on limits at infinity. So $y = 2$ is the only one.

Where it is defined. Let $0 < \abs{x - 3} < 1$, so that $2 < x < 3$ or $3 < x < 4$, by part (e) of the proposition on distance inequalities. Then $x + 1 > 3 > 0$ (adding 1, property 3 of the order rules), so $\sqrt{x + 1}$ is defined (the square-root remark); and $x^2 - 9 = (x - 3)(x + 3)$ is not 0, by part (a) of the sign rules, since $x \ne 3$, and $x \ne -3$ because $x > 2 > -3$ (transitivity, property 2 of the order rules, and then trichotomy, property 1). So the function is defined at every such $x$.

Rationalise and cancel. For such $x$, $\sqrt{x + 1} \ge 0$, so $\sqrt{x + 1} + 2 \ge 2 > 0$ (adding 2, property 3), and we may multiply the numerator and the denominator by this non-zero number, which we call $s$. Since $\bigl(\sqrt{x + 1}\bigr)^2 = x + 1$ (the square-root remark), and cancelling the non-zero factor $x - 3$,

The limit. All the functions below are defined at every point of $(2, 4)$. By part (a) of direct substitution, $x + 1 \to 4$ and $x + 3 \to 6$ as $x \to 3$. Since $4 > 0$, the square-root law, part (f) of the limit laws, gives $\sqrt{x + 1} \to \sqrt{4} = 2$ (as $2 \ge 0$ and $2^2 = 4$). By part (a) and the sum law, part (b), $s = \sqrt{x + 1} + 2 \to 4$; by the product law, part (c), the denominator $(x + 3)\,s$ approaches $6 \cdot 4 = 24$; and by part (a) and the quotient law, part (d), since $24 \ne 0$, the quotient approaches $\frac{1}{24}$. The original function agrees with it at every $x$ with $0 < \abs{x - 3} < 1$, so by the lemma on functions that agree except at a point, with $r = 1$,

Where it is defined. Let $0 < \abs{x} < \frac{\pi}{2}$. If $0 < x < \frac{\pi}{2}$, then $\cos x > 0$ by property 8 of the point $P(t)$. If $-\frac{\pi}{2} < x < 0$, multiplying by ${-1}$ reverses both inequalities (property 5(c) of the order rules), so $0 < -x < \frac{\pi}{2}$, and $\cos x = \cos(-x) > 0$ by properties 4 and 8 of $P(t)$. By part (c) of the lemma on sine and cosine near 0, $\cos x < \frac{\sin x}{x}$, so $\frac{\sin x}{x} > 0$ by transitivity (property 2 of the order rules). In particular $\sin x \ne 0$: $\sin x = \frac{\sin x}{x} \cdot x$ is a product of two non-zero numbers, so it is not 0, by part (a) of the sign rules. So the function $F(x) = \frac{x^2 \sin(1/x)}{\sin x}$ is defined at every such $x$, and there

The right-hand side is defined at every $x$ with $0 < \abs{x} < \frac{\pi}{2}$, since $\frac{\sin x}{x} \ne 0$ there.

The first factor. By the theorem on the limit of $\frac{\sin x}{x}$, $\frac{\sin x}{x} \to 1$, which is not 0, and $1 \to 1$ by part (a) of the limit laws. Both are defined at every $x \ne 0$, so the quotient law, part (d), gives $\frac{1}{\sin x / x} \to \frac11 = 1$.

The second factor. Let $x \ne 0$. By part (a) of the proposition on the bounds of sine and cosine, with $t = \frac{1}{x}$, $-1 \le \sin(1/x) \le 1$, which says $\abs{\sin(1/x)} \le 1$, by part (b) of the proposition on distance inequalities (centre 0, radius 1). By property 4 of the absolute value, and multiplying $\abs{\sin(1/x)} \le 1$ by the number $\abs{x} \ge 0$ (property 1 of the absolute value), which gives only $\le$ (property 5(b) of the order rules),

By part (a) of the lemma on limits at 0, $\lim_{x \to 0} \abs{x} = 0$. By the remark on squeezing the distance, with $a = 0$, $L = 0$, $r = 1$ and $b(x) = \abs{x}$, $\lim_{x \to 0} x\sin(1/x) = 0$.

Combine. By the product law, part (c) of the limit laws, the right-hand side above approaches $1 \cdot 0 = 0$. It equals $F(x)$ at every $x$ with $0 < \abs{x} < \frac{\pi}{2}$, so by the lemma on functions that agree except at a point, with $r = \frac{\pi}{2}$,

(a) On $[0, \infty)$, $P$ agrees with the rational function $\frac{p}{q}$, where $p(x) = x + 2000$ and $q(x) = x + 200$, both of degree 1 and leading coefficient 1. By part (c) of the proposition on rational functions at infinity, $\frac{p}{q}$ has the limit $\frac{1}{1} = 1$ as $x \to \infty$. Both are defined, and equal, at every $x > 0$, so by property 2 of the remark on limits at infinity (only the tail matters), with $c = 0$, $\lim_{x \to \infty} P(x) = 1$, in per cent: the 20 faulty parts made before the repair become a smaller and smaller share of all the parts, so the percentage approaches the 1 per cent of the repaired machine.

(b) Let $x \ge 0$. Adding 200 to $0 \le x$ gives $200 \le x + 200$ (property 3 of the order rules), so $x + 200 > 0$, since $0 < 200$ (transitivity, property 2). Multiplying by the positive number $x + 200$ keeps a strict inequality (property 5(a)), and multiplying back by its reciprocal $\frac{1}{x + 200}$, which is positive (property 6), keeps it too. So $P(x) < 2$ holds exactly when

Adding $-x - 400$ to both sides, or back $x + 400$ (property 3 both ways), this holds exactly when $1600 < x$.

So every $x > 1600$ has $P(x) < 2$: such an $x$ is positive, since $1600 > 0$ (transitivity, property 2), so the equivalence above applies to it. So $N = 1600$ works. No smaller $N$ does: if $N < 1600$, then $x = 1600$ satisfies $x > N$ and $x \ge 0$, but

which is not less than 2. So the smallest $N$ is 1600: the share of faulty parts is below 2 per cent exactly when the machine has made more than 1600 parts since the repair.

In the language of the definition of a limit at infinity: for $x \ge 0$, writing 1 as $\frac{x + 200}{x + 200}$ and using $(x + 2000) - (x + 200) = 1800$,

because a quotient of two positive numbers is positive (part (c) of the sign rules). So $\abs{P(x) - 1} = P(x) - 1$ (the definition of the absolute value), which is less than 1 exactly when $P(x) < 2$ (adding 1, or back -1, property 3). So 1600 is the smallest threshold that wins the round $\eps = 1$, and the share of faulty parts stays above 1 per cent throughout.

Where it is defined. For every real $x$, $x^2 \ge 0$ (the “in particular” of part (c) of the sign rules), so $4x^2 \ge 0$ (property 5(b) of the order rules) and $4x^2 + 1 \ge 1 > 0$ (adding 1, property 3). So the root is defined, and $f$ is defined at every $x \ne 1$. Moreover $1 \ge 0$, $\sqrt{4x^2 + 1} \ge 0$, and $1^2 = 1 \le 4x^2 + 1 = \bigl(\sqrt{4x^2 + 1}\bigr)^2$, so $\sqrt{4x^2 + 1} \ge 1$ by the order part of the square-root remark (and its uniqueness part for $=$).

(a) Let $x < 0$, so that $-x > 0$ (multiplying by ${-1}$, property 5(c)). Then $x^2 > 0$ (part (c) of the sign rules), so $\frac{1}{x^2} > 0$ (property 6) and $4 + \frac{1}{x^2} > 0$ (adding two positive numbers, property 4). Multiplying $\sqrt{4 + \frac{1}{x^2}} \ge 0$ (the square-root remark) by $-x \ge 0$ gives $-x\sqrt{4 + \frac{1}{x^2}} \ge 0$ (property 5(b)), and this number has the square $x^2\bigl(4 + \frac{1}{x^2}\bigr) = 4x^2 + 1$. By the uniqueness in the square-root remark it is $\sqrt{4x^2 + 1}$. Dividing the numerator and the denominator by the positive number $-x$, and using $\frac{x - 1}{-x} = -1 + \frac{1}{x}$,

As $x \to -\infty$, $\frac{1}{x} \to 0$ and $\frac{1}{x^2} \to 0$, by part (a) of the proposition on rational functions at infinity. By parts (a) and (b) of the limit laws at infinity, as $x \to -\infty$, $4 + \frac{1}{x^2} \to 4 > 0$, so the root approaches $\sqrt{4} = 2$ by part (f) (as $2 \ge 0$ and $2^2 = 4$, the square-root remark); and $-1 + \frac{1}{x} \to -1 \ne 0$. By the quotient law, part (d), the right-hand side approaches $\frac{2}{-1} = -2$. It equals $f(x)$ at every $x < 0$, so by property 2 of the remark on limits at infinity, $\lim_{x \to -\infty} f(x) = -2$.

(b) For $x \ne 1$, $f(x) = h(x)\,g(x)$ with $h(x) = \sqrt{4x^2 + 1}$ and $g(x) = \frac{1}{x - 1}$. By part (c) of the proposition on reciprocal powers, with $a = 1$ and $n = 1$, which is odd, $\lim_{x \to 1^{-}} g(x) = -\infty$. On $(0, 1)$ all three functions are defined and $h(x) \ge 1$, as shown above. By the left-hand version of part (a) of the proposition on a factor that stays away from 0, with $m = 1$ and the interval $(0, 1)$, $\lim_{x \to 1^{-}} f(x) = -\infty$.

(As $x \to \infty$, the same steps with $x$ in place of $-x$ give the limit 2: the graph has the two horizontal asymptotes $y = -2$ and $y = 2$, and the vertical asymptote $x = 1$.)

(a) Multiplying $L > 0$ by the positive number $\frac12$ keeps the inequality (property 5(a) of the order rules), so $\eps = \frac{L}{2}$ is positive. By the definition of the limit, there is a $\delta > 0$ such that $f$ is defined at every $x$ with $0 < \abs{x - a} < \delta$, and $\abs{f(x) - L} < \frac{L}{2}$ there. Let $r = \delta$, and let $0 < \abs{x - a} < r$. By part (a) of the proposition on distance inequalities, with centre $L$ and radius $\frac{L}{2}$,

and the left-hand inequality says $f(x) > \frac{L}{2}$.

(b) Each $f$ below is defined at every real $x$. If $x \ne a$, then $x - a \ne 0$, so $(x - a)^2 \ne 0$ by part (a) of the sign rules, and $F(x)$ is defined. We write $g(x) = \frac{1}{x - a}$ for $x \ne a$, the function of the proposition on reciprocal powers with $n = 1$, which is odd: by its parts (a) and (c), $\lim_{x \to a^{+}} g(x) = \infty$ and $\lim_{x \to a^{-}} g(x) = -\infty$.

(i) $f(x) = \abs{x - a}$. First, $f(x) \to 0$. Let $\eps > 0$ and $\delta = \eps$. If $0 < \abs{x - a} < \delta$, then $\abs{x - a} \ge 0$ (property 1 of the absolute value), so it is its own absolute value (the definition of the absolute value), and $\abs{f(x) - 0} = \abs{x - a} < \eps$. So $\lim_{x \to a} f(x) = 0$, by the definition of the limit.

Next, let $x \ne a$. Then $\abs{x - a} \ne 0$ (property 1), and $(x - a)^2 = \abs{x - a}^2$ (property 5), so cancelling $\abs{x - a}$ gives $F(x) = \frac{1}{\abs{x - a}}$. If $x > a$, then $x - a > 0$ (part (b) of the sign rules), so $\abs{x - a} = x - a$ (the definition of the absolute value) and $F(x) = 1 \cdot g(x)$. If $x < a$, then $x - a < 0$, so $\abs{x - a} = -(x - a)$ and $F(x) = (-1) \cdot g(x)$.

• From the right. By part (a) of the proposition on a factor that stays away from 0, with $h(x) = 1 \ge 1 = m$ on $(a, a + 1)$, and since $g(x) \to \infty$ as $x \to a^{+}$, $\lim_{x \to a^{+}} F(x) = \infty$.

• From the left. By the left-hand version of part (b) of that proposition, with $h(x) = -1 \le -1 = -m$ on $(a - 1, a)$, and since $g(x) \to -\infty$ as $x \to a^{-}$, $\lim_{x \to a^{-}} F(x) = \infty$.

$F$ is defined at every point of $(a - 1, a)$ and of $(a, a + 1)$, so $\lim_{x \to a} F(x) = \infty$, by part (c) of the definition of infinite limits.

(ii) $f(x) = (x - a)^2$. This is a polynomial with $f(a) = 0$, so $\lim_{x \to a} f(x) = 0$ by part (a) of direct substitution. At every $x \ne a$, $F(x) = 1$, a non-zero number divided by itself. The constant function 1 has the limit 1 (part (a) of the limit laws), and it equals $F$ at every $x$ with $0 < \abs{x - a} < 1$, so $\lim_{x \to a} F(x) = 1$, by the lemma on functions that agree except at a point, with $r = 1$.

(iii) $f(x) = x - a$. This is a polynomial with $f(a) = 0$, so $\lim_{x \to a} f(x) = 0$, by part (a) of direct substitution. At every $x \ne a$, cancelling the non-zero factor $x - a$ gives $F(x) = \frac{1}{x - a} = g(x)$, so $F$ is the function $g$.

• No real limit. The right-hand limit of $F$ at $a$ is $\infty$, so $\lim_{x \to a} F(x)$ does not exist, by part (b) of the proposition that an infinite limit is not a real limit, with $r = 1$.

• Not $\infty$. If $\lim_{x \to a} F(x) = \infty$, then $\lim_{x \to a^{-}} F(x) = \infty$, by part (c) of the definition of infinite limits. With the height $M = 1$, the precise form, part (b), gives a $\delta > 0$ with $F(x) > 1$ at every $x$ in $(a - \delta, a)$. Take such an $x$. Then $x - a < 0$ (part (b) of the sign rules), so $F(x) = \frac{1}{x - a} < 0$ (part (c) of the sign rules), and $F(x) < 0 < 1$ gives $F(x) < 1$ (transitivity, property 2 of the order rules). Both $F(x) > 1$ and $F(x) < 1$ cannot hold (trichotomy, property 1).

• Not $-\infty$. In the same way, $\lim_{x \to a} F(x) = -\infty$ would give $\lim_{x \to a^{+}} F(x) = -\infty$, and so an $x > a$ with $F(x) < -1$. But there $x - a > 0$, so $F(x) = \frac{1}{x - a} > 0$ (property 6 of the order rules), and $-1 < 0 < F(x)$ gives $F(x) > -1$: again impossible.

So when $L = 0$ the quotient may tend to $\infty$, to a real number, or to nothing at all: the conclusion for $L > 0$ rests on that hypothesis.

Let $F(x) = \sqrt{x^2 + ax + 1} - bx$, and let $x > \abs{a}$.

Where it is defined. Since $\abs{a} \ge 0$ (property 1 of the absolute value), $x > 0$ by transitivity (property 2 of the order rules). By property 3 of the absolute value, $a \ge -\abs{a}$; adding this to $x > \abs{a}$ (property 4 of the order rules) gives $x + a > 0$. Multiplying by the positive number $x$ (property 5(a)), $x(x + a) > 0$, so $x^2 + ax + 1 > 1 > 0$ (adding 1, property 3). So $F$ is defined at every $x > \abs{a}$, and so is

which is positive, as a quotient of two positive numbers (part (c) of the sign rules). Multiplying $\sqrt{1 + \frac{a}{x} + \frac{1}{x^2}} \ge 0$ (the square-root remark) by $x \ge 0$ gives $x\sqrt{1 + \frac{a}{x} + \frac{1}{x^2}} \ge 0$ (property 5(b) of the order rules). Its square is $x^2 + ax + 1$, so by the uniqueness in the square-root remark,

The root, after dividing by $x$. By part (a) of the proposition on rational functions at infinity, $\frac{1}{x} \to 0$ and $\frac{1}{x^2} \to 0$ as $x \to \infty$. By parts (a), (b) and (c) of the limit laws at infinity, $1 + \frac{a}{x} + \frac{1}{x^2} \to 1$, and since $1 > 0$, part (f) gives

as $1 \ge 0$ and $1^2 = 1$ (the square-root remark).

$b$ must be 1. Suppose that $\lim_{x \to \infty} F(x) = -2$. For $x > \abs{a}$, dividing by $x > 0$,

By the difference law, part (b), the right-hand side approaches $1 - b$. On the other hand, $\frac{F(x)}{x} = F(x) \cdot \frac{1}{x}$ approaches $({-2}) \cdot 0 = 0$, by the product law, part (c). By uniqueness, property 1 of the remark on limits at infinity, $1 - b = 0$, so $b = 1$.

$a$ must be ${-4}$. With $b = 1$, and $x > \abs{a}$, the number $\sqrt{x^2 + ax + 1} + x$ is positive: adding $0 \le \sqrt{x^2 + ax + 1}$ and $0 < x$ gives $0 < \sqrt{x^2 + ax + 1} + x$ (property 4 of the order rules). Multiplying and dividing by it, using $\bigl(\sqrt{x^2 + ax + 1}\bigr)^2 = x^2 + ax + 1$ (the square-root remark), and then dividing the numerator and the denominator by $x > 0$,

The numerator approaches $a$ (parts (a) and (b)), and the denominator $1 + 1 = 2 \ne 0$, so by the quotient law, part (d), $F(x) \to \frac{a}{2}$. By uniqueness again, $\frac{a}{2} = -2$, so $a = -4$ (multiplying by 2).

The pair works. Conversely, for $a = -4$ and $b = 1$, the last step shows that $\lim_{x \to \infty} F(x) = \frac{-4}{2} = -2$. So $a = -4$ and $b = 1$ is a pair with the limit ${-2}$, and the steps above show that it is the only one: for large $x$ the graph of $\sqrt{x^2 - 4x + 1}$ approaches the line $y = x - 2$.