Limits, Worked Solutions

Ques­tion 1: Lim­its by sub­sti­tu­tion and by fac­toris­ing#

The prob­lem#

Eval­u­ate:

(a) $\displaystyle \lim_{x \to -1}(2x^2 - 3x + 4)$

(b) $\displaystyle \lim_{x \to 4} \frac{x^2 - 16}{x - 4}$

(c) $\displaystyle \lim_{x \to 2} \frac{x^2 - 5x + 6}{x^2 - 4}$

Under­stand­ing the prob­lem#

Part (a) is a poly­no­mial. Parts (b) and (c) are ratio­nal func­tions (a poly­no­mial divided by a poly­no­mial). We must check what hap­pens when we sub­sti­tute the value of $x$.

The idea#

For a poly­no­mial, $\displaystyle \lim_{x \to a} p(x) = p(a)$. For a ratio­nal func­tion, sub­sti­tute if the denom­i­na­tor is not zero. If sub­sti­tu­tion gives $\displaystyle \tfrac{0}{0}$, the numer­a­tor and denom­i­na­tor share the fac­tor $(x - a)$; can­cel it (allowed because $x \ne a$ while $x$ approaches $a$) and then sub­sti­tute.

Step-by-step solu­tion#

Part (a)

Step 1. This is a poly­no­mial, so sub­sti­tute $x = -1$ directly.

$$2(-1)^2 - 3(-1) + 4 = 2 + 3 + 4 = 9$$

Part (b)

Step 1. Try sub­sti­tu­tion: numer­a­tor $16 - 16 = 0$, denom­i­na­tor $4 - 4 = 0$. This is $\displaystyle \tfrac{0}{0}$, so we must sim­plify.

Step 2. Fac­torise the numer­a­tor as a dif­fer­ence of squares and can­cel $(x - 4)$.

$$\displaystyle \frac{x^2 - 16}{x - 4} = \frac{(x - 4)(x + 4)}{x - 4} = x + 4 \qquad (x \ne 4)$$

Step 3. Now sub­sti­tute.

$$\displaystyle \lim_{x \to 4}(x + 4) = 8$$

Part (c)

Step 1. Try sub­sti­tu­tion: numer­a­tor $4 - 10 + 6 = 0$, denom­i­na­tor $4 - 4 = 0$. Again $\displaystyle \tfrac{0}{0}$.

Step 2. Fac­torise both. For $x^2 - 5x + 6$ we need two num­bers with prod­uct $6$ and sum $-5$: they are $-2$ and $-3$.

$$\displaystyle \frac{x^2 - 5x + 6}{x^2 - 4} = \frac{(x - 2)(x - 3)}{(x - 2)(x + 2)} = \frac{x - 3}{x + 2} \qquad (x \ne 2)$$

Step 3. Sub­sti­tute $x = 2$.

$$\displaystyle \lim_{x \to 2}\frac{x - 3}{x + 2} = \frac{2 - 3}{2 + 2} = -\frac{1}{4}$$

Check­ing the answer#

For (b), try $x = 4.01$: $\displaystyle \frac{16.0801 - 16}{0.01} = 8.01$, close to $8$. For (c), try $x = 2.01$: $\displaystyle \frac{-0.99}{4.01} \approx -0.247$, close to $-0.25$.

Answer#

(a) $9$ (b) $8$ (c) $\displaystyle -\frac{1}{4}$

Com­mon mis­take to avoid#

Writ­ing $\displaystyle \tfrac{0}{0} = 0$ or "not defined, so no limit". $\displaystyle \tfrac{0}{0}$ only tells you to sim­plify first.

Ques­tion 2: The stan­dard limit $\displaystyle \frac{x^n - a^n}{x - a}$#

The prob­lem#

Eval­u­ate:

(a) $\displaystyle \lim_{x \to 1} \frac{x^7 - 1}{x - 1}$

(b) $\displaystyle \lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}$

(c) $\displaystyle \lim_{x \to 0} \frac{(1 + x)^6 - 1}{x}$

Under­stand­ing the prob­lem#

All three give $\displaystyle \tfrac{0}{0}$ on sub­sti­tu­tion. Each one can be matched with the pat­tern $\displaystyle \frac{x^n - a^n}{x - a}$.

The idea#

Use the stan­dard result

$$\displaystyle \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}.$$

Step-by-step solu­tion#

Part (a)

Step 1. Write $1 = 1^7$, so the expres­sion is $\displaystyle \frac{x^7 - 1^7}{x - 1}$ with $n = 7$, $a = 1$.

Step 2. Apply the result.

$$\displaystyle \lim_{x \to 1} \frac{x^7 - 1^7}{x - 1} = 7 \cdot 1^{6} = 7$$

Part (b)

Step 1. Here $81 = 3^4$ and $9 = 3^2$. The denom­i­na­tor is not $x - 3$, so divide both the numer­a­tor and the denom­i­na­tor by $(x - 3)$.

$$\displaystyle \frac{x^4 - 81}{x^2 - 9} = \frac{\dfrac{x^4 - 3^4}{x - 3}}{\dfrac{x^2 - 3^2}{x - 3}}$$

Step 2. Take the limit of the top and bot­tom sep­a­rately using the stan­dard result.

$$\displaystyle \begin{aligned} \lim_{x \to 3}\frac{x^4 - 3^4}{x - 3} &= 4 \cdot 3^3 = 108\\ \lim_{x \to 3}\frac{x^2 - 3^2}{x - 3} &= 2 \cdot 3 = 6 \end{aligned}$$

Step 3. The bot­tom limit is not zero, so we may divide.

$$\displaystyle \lim_{x \to 3}\frac{x^4 - 81}{x^2 - 9} = \frac{108}{6} = 18$$

Part (c)

Step 1. Put $y = 1 + x$. As $x \to 0$, $y \to 1$, and $x = y - 1$.

$$\displaystyle \lim_{x \to 0} \frac{(1 + x)^6 - 1}{x} = \lim_{y \to 1} \frac{y^6 - 1^6}{y - 1}$$

Step 2. Apply the result with $n = 6$, $a = 1$.

$$= 6 \cdot 1^5 = 6$$

Check­ing the answer#

For (b) you can also fac­torise: $\displaystyle \frac{(x^2 - 9)(x^2 + 9)}{x^2 - 9} = x^2 + 9 \to 18$. Same answer.

Answer#

(a) $7$ (b) $18$ (c) $6$

Ques­tion 3: Trigono­met­ric lim­its#

The prob­lem#

Eval­u­ate:

(a) $\displaystyle \lim_{x \to 0} \frac{\sin 4x}{x}$

(b) $\displaystyle \lim_{x \to 0} \frac{\tan 3x}{\sin 2x}$

(c) $\displaystyle \lim_{x \to 0} \frac{1 - \cos 4x}{x^2}$

Under­stand­ing the prob­lem#

Each gives $\displaystyle \tfrac{0}{0}$ at $x = 0$. The key is to reshape each expres­sion so that $\displaystyle \frac{\sin \theta}{\theta}$ or $\displaystyle \frac{\tan \theta}{\theta}$ appears with the same angle on top and bot­tom.

The idea#

Use $\displaystyle \lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$ and $\displaystyle \lim_{\theta \to 0}\frac{\tan \theta}{\theta} = 1$. As $x \to 0$, any $\theta = kx$ also tends to $0$.

Step-by-step solu­tion#

Part (a)

Step 1. Mul­ti­ply and divide by $4$ so that the angle $4x$ also appears in the denom­i­na­tor.

$$\displaystyle \frac{\sin 4x}{x} = 4 \cdot \frac{\sin 4x}{4x}$$

Step 2. As $x \to 0$, $4x \to 0$, so $\displaystyle \frac{\sin 4x}{4x} \to 1$.

$$\displaystyle \lim_{x \to 0}\frac{\sin 4x}{x} = 4 \times 1 = 4$$

Part (b)

Step 1. Arrange the expres­sion into stan­dard pieces.

$$\displaystyle \frac{\tan 3x}{\sin 2x} = \frac{\tan 3x}{3x} \cdot \frac{2x}{\sin 2x} \cdot \frac{3x}{2x}$$

Check: the $3x$ and $2x$ you intro­duced can­cel out, so noth­ing has changed.

Step 2. Take lim­its of each piece.

$$\displaystyle \lim_{x \to 0}\frac{\tan 3x}{\sin 2x} = 1 \times 1 \times \frac{3}{2} = \frac{3}{2}$$

Part (c)

Step 1. Use the iden­tity $1 - \cos 2\theta = 2\sin^2 \theta$ with $\theta = 2x$.

$$1 - \cos 4x = 2\sin^2 2x$$

Step 2. Rewrite and make the angle $2x$ appear in the denom­i­na­tor.

$$\displaystyle \frac{1 - \cos 4x}{x^2} = \frac{2\sin^2 2x}{x^2} = 2 \cdot 4 \cdot \left(\frac{\sin 2x}{2x}\right)^2$$

(Because $\displaystyle x^2 = \frac{(2x)^2}{4}$, divid­ing by $x^2$ is mul­ti­ply­ing by $\displaystyle \frac{4}{(2x)^2}$.)

Step 3. Take the limit.

$$\displaystyle \lim_{x \to 0}\frac{1 - \cos 4x}{x^2} = 8 \times 1^2 = 8$$

Check­ing the answer#

Try $x = 0.01$ in (a): $\sin 0.04 \approx 0.039989$, divided by $0.01$ gives $3.9989$, close to $4$.

Answer#

(a) $4$ (b) $\displaystyle \frac{3}{2}$ (c) $8$

Com­mon mis­take to avoid#

Can­celling "sin" or can­celling $x$ inside $\sin 4x$. $\displaystyle \frac{\sin 4x}{x}$ is not $\sin 4$. Always build $\displaystyle \frac{\sin(\text{angle})}{\text{same angle}}$.

Ques­tion 4: A root and a shifted angle#

The prob­lem#

Eval­u­ate:

(a) $\displaystyle \lim_{x \to 0} \frac{\sqrt{4 + x} - 2}{x}$

(b) $\displaystyle \lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi(\pi - x)}$

Under­stand­ing the prob­lem#

Both give $\displaystyle \tfrac{0}{0}$. In (a) there is a square root, which is a power $\displaystyle \tfrac{1}{2}$. In (b), $x$ approaches $\pi$, not $0$, so the stan­dard $\displaystyle \frac{\sin \theta}{\theta}$ result can­not be used until we change the vari­able.

The idea#

(a) Sub­sti­tute $y = 4 + x$ and use $\displaystyle \frac{y^n - a^n}{y - a} \to n a^{n-1}$ with $\displaystyle n = \tfrac{1}{2}$ (the les­son says the result holds for ratio­nal $n$ when $a > 0$).
(b) Sub­sti­tute $h = \pi - x$, which tends to $0$, and use $\displaystyle \frac{\sin h}{h} \to 1$.

Step-by-step solu­tion#

Part (a)

Step 1. Let $y = 4 + x$. Then $x = y - 4$, $\sqrt{4 + x} = y^{1/2}$, $2 = 4^{1/2}$, and as $x \to 0$, $y \to 4$.

$$\displaystyle \lim_{x \to 0} \frac{\sqrt{4 + x} - 2}{x} = \lim_{y \to 4} \frac{y^{1/2} - 4^{1/2}}{y - 4}$$

Step 2. Apply the stan­dard result with $\displaystyle n = \tfrac{1}{2}$, $a = 4$.

$$\displaystyle = \frac{1}{2}\cdot 4^{-1/2} = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$$

Part (b)

Step 1. Let $h = \pi - x$. As $x \to \pi$, $h \to 0$.

$$\displaystyle \lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi(\pi - x)} = \lim_{h \to 0} \frac{\sin h}{\pi h} = \frac{1}{\pi}\lim_{h \to 0}\frac{\sin h}{h}$$

Step 2. Use $\displaystyle \frac{\sin h}{h} \to 1$.

$$\displaystyle = \frac{1}{\pi} \times 1 = \frac{1}{\pi}$$

Check­ing the answer#

For (a), put $x = 0.01$: $\sqrt{4.01} \approx 2.0025$, so the frac­tion is about $\displaystyle \frac{0.0025}{0.01} = 0.25$.

Answer#

(a) $\displaystyle \frac{1}{4}$ (b) $\displaystyle \frac{1}{\pi}$

Ques­tion 5: Left and right lim­its of a piece­wise func­tion#

The prob­lem#

Find the left and right lim­its at $0$ of

$$f(x) = \begin{cases} 2x + 3, & x \le 0 \\ 3(x + 1), & x > 0 \end{cases}$$

Does the limit exist?

Under­stand­ing the prob­lem#

The func­tion has one for­mula to the left of $0$ (and at $0$) and another to the right. The left limit uses $x < 0$, the right limit uses $x > 0$.

The idea#

Com­pute each one-sided limit using the for­mula valid on that side. The limit exists exactly when the two are equal.

Step-by-step solu­tion#

Step 1. Left limit: for $x < 0$, $f(x) = 2x + 3$.

$$\displaystyle \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-}(2x + 3) = 3$$

Step 2. Right limit: for $x > 0$, $f(x) = 3(x + 1)$.

$$\displaystyle \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 3(x + 1) = 3$$

Step 3. Com­pare: both equal $3$, so the limit exists.

Check­ing the answer#

$f(-0.01) = 2.98$ and $f(0.01) = 3.03$; both are close to $3$.

Answer#

Left limit $3$, right limit $3$; yes, $\displaystyle \lim_{x \to 0} f(x) = 3$.

Ques­tion 6: Choos­ing $k$ so that the limit exists#

The prob­lem#

For what $k$ does $\displaystyle \lim_{x \to 2} f(x)$ exist if $f(x) = kx - 1$ for $x < 2$ and $f(x) = x^2 + 1$ for $x \ge 2$?

Under­stand­ing the prob­lem#

$k$ is an unknown con­stant. The limit at $2$ exists only if the left and right lim­its at $2$ are equal. That gives an equa­tion for $k$.

The idea#

Find each one-sided limit in terms of $k$, then set them equal.

Step-by-step solu­tion#

Step 1. Left limit (use $kx - 1$).

$$\displaystyle \lim_{x \to 2^-}(kx - 1) = 2k - 1$$

Step 2. Right limit (use $x^2 + 1$).

$$\displaystyle \lim_{x \to 2^+}(x^2 + 1) = 5$$

Step 3. Set them equal and solve.

$$2k - 1 = 5 \quad\Rightarrow\quad 2k = 6 \quad\Rightarrow\quad k = 3$$

Check­ing the answer#

With $k = 3$, the left piece is $3x - 1$, which gives $5$ at $x = 2$, match­ing $2^2 + 1 = 5$.

Answer#

$k = 3$.

Ques­tion 7: The great­est inte­ger func­tion#

The prob­lem#

Does $\displaystyle \lim_{x \to 0} [x]$ exist? What about $\displaystyle \lim_{x \to 0.5} [x]$?

Under­stand­ing the prob­lem#

$[x]$ is the great­est inte­ger less than or equal to $x$. For exam­ple $[0.3] = 0$, $[-0.3] = -1$. Its graph is a stair­case that jumps at every inte­ger.

The idea#

Near an inte­ger the func­tion jumps, so check the left and right lim­its sep­a­rately. Near a non-inte­ger the func­tion is con­stant, so the limit is just that con­stant.

Step-by-step solu­tion#

Part (a): at $x = 0$

Step 1. For $-1 \le x < 0$, $[x] = -1$. So the left limit is $-1$.

Step 2. For $0 \le x < 1$, $[x] = 0$. So the right limit is $0$.

Step 3. $-1 \ne 0$, so the limit does not exist.

Part (b): at $x = 0.5$

Step 1. For all $x$ between $0$ and $1$ (on both sides of $0.5$), $[x] = 0$.

Step 2. So both one-sided lim­its are $0$, and $\displaystyle \lim_{x \to 0.5}[x] = 0$.

Check­ing the answer#

$[-0.001] = -1$ while $[0.001] = 0$: the val­ues really do split at $0$. $[0.499] = [0.501] = 0$.

Answer#

At $0$: no (left limit $-1$, right limit $0$). At $0.5$: yes, the limit is $0$.

Ques­tion 8: Divid­ing through by $x$#

The prob­lem#

Eval­u­ate $\displaystyle \lim_{x \to 0} \frac{x\cos x + \sin x}{x^2 + \tan x}$.

Under­stand­ing the prob­lem#

At $x = 0$ the numer­a­tor is $0 + 0 = 0$ and the denom­i­na­tor is $0 + 0 = 0$: the form $\displaystyle \tfrac{0}{0}$.

The idea#

Divide every term of the numer­a­tor and the denom­i­na­tor by $x$. This pro­duces $\displaystyle \frac{\sin x}{x}$ and $\displaystyle \frac{\tan x}{x}$, whose lim­its we know.

Step-by-step solu­tion#

Step 1. Divide top and bot­tom by $x$ (allowed since $x \ne 0$).

$$\displaystyle \frac{x\cos x + \sin x}{x^2 + \tan x} = \frac{\cos x + \dfrac{\sin x}{x}}{x + \dfrac{\tan x}{x}}$$

Step 2. Take the limit of each piece.

$$\begin{aligned} \text{numerator} &\to \cos 0 + 1 = 1 + 1 = 2\\ \text{denominator} &\to 0 + 1 = 1 \end{aligned}$$

Step 3. The denom­i­na­tor limit is not zero, so divide.

$$\displaystyle \lim_{x \to 0} \frac{x\cos x + \sin x}{x^2 + \tan x} = \frac{2}{1} = 2$$

Check­ing the answer#

At $x = 0.01$: numer­a­tor $\approx 0.0099995 + 0.0099998 \approx 0.0200$, denom­i­na­tor $\approx 0.0001 + 0.0100003 \approx 0.0101$; ratio $\approx 1.98$, close to $2$.

Answer#

$2$

Ques­tion 9: $\displaystyle \frac{\sin x - \sin a}{x - a}$#

The prob­lem#

Eval­u­ate $\displaystyle \lim_{x \to a} \frac{\sin x - \sin a}{x - a}$.

Under­stand­ing the prob­lem#

At $x = a$ both top and bot­tom are $0$. $a$ is a fixed num­ber; the answer will be in terms of $a$.

The idea#

Turn the dif­fer­ence of sines into a prod­uct with the sum-to-prod­uct for­mula

$$\displaystyle \sin C - \sin D = 2\cos\frac{C + D}{2}\sin\frac{C - D}{2},$$

then cre­ate $\displaystyle \frac{\sin h}{h}$ with $\displaystyle h = \frac{x - a}{2}$.

Step-by-step solu­tion#

Step 1. Apply the for­mula with $C = x$, $D = a$.

$$\displaystyle \frac{\sin x - \sin a}{x - a} = \frac{2\cos\dfrac{x + a}{2}\sin\dfrac{x - a}{2}}{x - a}$$

Step 2. Write $\displaystyle x - a = 2 \cdot \frac{x - a}{2}$ so the $2$ can­cels.

$$\displaystyle = \cos\frac{x + a}{2} \cdot \frac{\sin\dfrac{x - a}{2}}{\dfrac{x - a}{2}}$$

Step 3. As $x \to a$, $\displaystyle \frac{x - a}{2} \to 0$, so the sec­ond fac­tor tends to $1$; the first tends to $\displaystyle \cos\frac{2a}{2} = \cos a$.

$$\displaystyle \lim_{x \to a} \frac{\sin x - \sin a}{x - a} = \cos a \times 1 = \cos a$$

Check­ing the answer#

Spe­cial case $a = 0$: the limit becomes $\displaystyle \frac{\sin x}{x} \to 1$, and $\cos 0 = 1$. It agrees.

Answer#

$\cos a$

Ques­tion 10: A limit with two con­stants#

The prob­lem#

Eval­u­ate $\displaystyle \lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}$ for non-zero $a, b$ with $a + b \ne 0$.

Under­stand­ing the prob­lem#

$a$ and $b$ are fixed non-zero num­bers. At $x = 0$ the form is $\displaystyle \tfrac{0}{0}$. The con­di­tion $a + b \ne 0$ will make sure we do not divide by zero at the end.

The idea#

Divide every term by $x$ to cre­ate $\displaystyle \frac{\sin ax}{x}$ and $\displaystyle \frac{\sin bx}{x}$, and use $\displaystyle \frac{\sin kx}{x} = k \cdot \frac{\sin kx}{kx} \to k$.

Step-by-step solu­tion#

Step 1. Divide top and bot­tom by $x$.

$$\displaystyle \frac{\sin ax + bx}{ax + \sin bx} = \frac{\dfrac{\sin ax}{x} + b}{a + \dfrac{\sin bx}{x}}$$

Step 2. Find the lim­its of the sine pieces.

$$\displaystyle \frac{\sin ax}{x} = a\cdot\frac{\sin ax}{ax} \to a, \qquad \frac{\sin bx}{x} = b\cdot\frac{\sin bx}{bx} \to b$$

Step 3. Sub­sti­tute these lim­its.

$$\displaystyle \lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx} = \frac{a + b}{a + b} = 1$$

This divi­sion is allowed because $a + b \ne 0$.

Check­ing the answer#

Take $a = 2$, $b = 3$, $x = 0.01$: top $\approx 0.019999 + 0.03 = 0.049999$, bot­tom $\approx 0.02 + 0.029996 = 0.049996$; ratio $\approx 1.0001$.

Answer#

$1$