Limits and Derivatives: Mixed Practice, Worked Solutions

Ques­tion 1: A limit by fac­toris­ing#

The prob­lem#

Eval­u­ate $\displaystyle \lim_{x \to 5} \frac{x^2 - 7x + 10}{x - 5}$.

Under­stand­ing the prob­lem#

You want the value the frac­tion approaches as $x$ gets close to $5$. The frac­tion itself is not defined at $x = 5$, but that does not mat­ter: a limit only looks at val­ues near $5$.

The idea#

Sub­sti­tut­ing $x = 5$ gives $\displaystyle \tfrac{25 - 35 + 10}{0} = \tfrac{0}{0}$. Since the numer­a­tor is $0$ at $x = 5$, $(x - 5)$ must be a fac­tor of it. Fac­torise and can­cel.

Step-by-step solu­tion#

Step 1. Check by sub­sti­tu­tion.

$$\displaystyle \frac{5^2 - 7(5) + 10}{5 - 5} = \frac{0}{0} \quad (\text{indeterminate}).$$

Step 2. Fac­torise the numer­a­tor: two num­bers with prod­uct $10$ and sum $-7$ are $-5$ and $-2$.

$$x^2 - 7x + 10 = (x - 5)(x - 2).$$

Step 3. Can­cel $(x - 5)$, which is allowed because $x \ne 5$ while $x$ is only approach­ing $5$.

$$\displaystyle \lim_{x \to 5}\frac{(x - 5)(x - 2)}{x - 5} = \lim_{x \to 5}(x - 2).$$

Step 4. Now sub­sti­tute.

$$\displaystyle \lim_{x \to 5}(x - 2) = 5 - 2 = 3.$$

Check­ing the answer#

Try $x = 5.01$: $\displaystyle \tfrac{25.1001 - 35.07 + 10}{0.01} = \tfrac{0.0301}{0.01} = 3.01$, very close to $3$. ✓

Answer#

$3$

Ques­tion 2: Using the stan­dard limit $\displaystyle \tfrac{x^n - a^n}{x - a}$#

The prob­lem#

Eval­u­ate $\displaystyle \lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}$.

Under­stand­ing the prob­lem#

Sub­sti­tu­tion gives $\displaystyle \tfrac{1 - 1}{1 - 1} = \tfrac{0}{0}$. Fac­toris­ing $x^{15} - 1$ by hand would be painful, so you need a smarter method.

The idea#

Both top and bot­tom look like $x^n - a^n$ with $a = 1$ (since $1^{15} = 1^{10} = 1$). Divide top and bot­tom by $(x - 1)$ so that each becomes the stan­dard limit $\displaystyle \lim_{x \to a}\tfrac{x^n - a^n}{x - a} = na^{n-1}$.

Step-by-step solu­tion#

Step 1. Divide numer­a­tor and denom­i­na­tor by $(x - 1)$, this does not change the frac­tion.

$$\displaystyle \frac{x^{15} - 1}{x^{10} - 1} = \frac{\dfrac{x^{15} - 1^{15}}{x - 1}}{\dfrac{x^{10} - 1^{10}}{x - 1}}.$$

Step 2. Take the limit of the top and bot­tom sep­a­rately using $na^{n-1}$ with $a = 1$.

$$\displaystyle \lim_{x \to 1}\frac{x^{15} - 1^{15}}{x - 1} = 15 \cdot 1^{14} = 15, \qquad \lim_{x \to 1}\frac{x^{10} - 1^{10}}{x - 1} = 10 \cdot 1^{9} = 10.$$

Step 3. Divide (allowed because the bot­tom limit, $10$, is not zero).

$$\displaystyle \lim_{x \to 1}\frac{x^{15} - 1}{x^{10} - 1} = \frac{15}{10} = \frac{3}{2}.$$

Check­ing the answer#

At $x = 1.001$: $x^{15} - 1 \approx 0.01511$ and $x^{10} - 1 \approx 0.01005$, ratio $\approx 1.503$. ✓

Answer#

$\displaystyle \frac{3}{2}$

Ques­tion 3: A trigono­met­ric limit with $\sin$ and $\tan$#

The prob­lem#

Eval­u­ate $\displaystyle \lim_{x \to 0} \frac{\sin 7x}{\tan 2x}$.

Under­stand­ing the prob­lem#

At $x = 0$, $\sin 0 = 0$ and $\tan 0 = 0$, so you get $\displaystyle \tfrac{0}{0}$. You need the stan­dard trigono­met­ric lim­its.

The idea#

Use $\displaystyle \lim_{\theta \to 0}\tfrac{\sin\theta}{\theta} = 1$ and $\displaystyle \lim_{\theta \to 0}\tfrac{\tan\theta}{\theta} = 1$. To use them, the angle inside must match the denom­i­na­tor: $\sin 7x$ needs $7x$ beneath it, and $\tan 2x$ needs $2x$. Mul­ti­ply and divide to cre­ate these.

Step-by-step solu­tion#

Step 1. Rewrite the frac­tion by mul­ti­ply­ing and divid­ing by $7x$ and $2x$.

$$\displaystyle \frac{\sin 7x}{\tan 2x} = \frac{\sin 7x}{7x} \cdot \frac{2x}{\tan 2x} \cdot \frac{7x}{2x}.$$

Step 2. Sim­plify the last fac­tor: $\displaystyle \tfrac{7x}{2x} = \tfrac{7}{2}$ (since $x \ne 0$).

Step 3. As $x \to 0$, $7x \to 0$ and $2x \to 0$, so

$$\displaystyle \frac{\sin 7x}{7x} \to 1, \qquad \frac{2x}{\tan 2x} \to 1.$$

Step 4. Mul­ti­ply the lim­its.

$$\displaystyle \lim_{x \to 0}\frac{\sin 7x}{\tan 2x} = 1 \cdot 1 \cdot \frac{7}{2} = \frac{7}{2}.$$

Check­ing the answer#

At $x = 0.01$: $\sin 0.07 \approx 0.06994$, $\tan 0.02 \approx 0.02000$, ratio $\approx 3.497$. ✓

Answer#

$\displaystyle \frac{7}{2}$

Ques­tion 4: Dif­fer­ence of cosines over $x^2$#

The prob­lem#

Eval­u­ate $\displaystyle \lim_{x \to 0} \frac{\cos 3x - \cos x}{x^2}$.

Under­stand­ing the prob­lem#

At $x = 0$ the top is $1 - 1 = 0$ and the bot­tom is $0$: again $\displaystyle \tfrac{0}{0}$. A dif­fer­ence of two cosines is awk­ward; a prod­uct of sines is easy to han­dle with $\displaystyle \tfrac{\sin\theta}{\theta}$.

The idea#

Use the iden­tity $\displaystyle \cos C - \cos D = -2\sin\frac{C + D}{2}\sin\frac{C - D}{2}$ to turn the dif­fer­ence into a prod­uct, then use $\displaystyle \lim\tfrac{\sin\theta}{\theta} = 1$ twice.

Step-by-step solu­tion#

Step 1. Apply the iden­tity with $C = 3x$, $D = x$.

$$\displaystyle \cos 3x - \cos x = -2\sin\frac{3x + x}{2}\sin\frac{3x - x}{2} = -2\sin 2x \sin x.$$

Step 2. Sub­sti­tute into the limit.

$$\displaystyle \lim_{x \to 0}\frac{-2\sin 2x\sin x}{x^2}.$$

Step 3. Split $x^2 = x \cdot x$, and make $\sin 2x$ sit over $2x$.

$$\displaystyle -2 \cdot \frac{\sin 2x}{x} \cdot \frac{\sin x}{x} = -2 \cdot 2 \cdot \frac{\sin 2x}{2x} \cdot \frac{\sin x}{x}.$$

Step 4. Take the limit: each frac­tion tends to $1$.

$$-2 \cdot 2 \cdot 1 \cdot 1 = -4.$$

Check­ing the answer#

The answer is neg­a­tive. That makes sense: for small $x \ne 0$, $3x$ is far­ther from $0$ than $x$, so $\cos 3x < \cos x$ and the numer­a­tor is neg­a­tive, while $x^2 > 0$.

Answer#

$-4$

Ques­tion 5: A limit at $\displaystyle \tfrac{\pi}{2}$ by chang­ing the vari­able#

The prob­lem#

Eval­u­ate $\displaystyle \lim_{x \to \pi/2} \frac{\tan 2x}{x - \frac{\pi}{2}}$.

Under­stand­ing the prob­lem#

At $\displaystyle x = \tfrac{\pi}{2}$, $\tan 2x = \tan\pi = 0$ and the denom­i­na­tor is $0$: $\displaystyle \tfrac{0}{0}$. The stan­dard lim­its work only as the vari­able tends to $0$, but here $\displaystyle x \to \tfrac{\pi}{2}$.

The idea#

Shift the vari­able so that it tends to $0$: put $\displaystyle y = x - \tfrac{\pi}{2}$. Then $y \to 0$ as $\displaystyle x \to \tfrac{\pi}{2}$. Use the fact that $\tan$ repeats every $\pi$.

Step-by-step solu­tion#

Step 1. Sub­sti­tute $\displaystyle y = x - \tfrac{\pi}{2}$, so $\displaystyle x = y + \tfrac{\pi}{2}$ and $2x = 2y + \pi$.

$$\displaystyle \lim_{x \to \pi/2}\frac{\tan 2x}{x - \frac{\pi}{2}} = \lim_{y \to 0}\frac{\tan(2y + \pi)}{y}.$$

Step 2. Since $\tan$ has period $\pi$, $\tan(2y + \pi) = \tan 2y$.

$$\displaystyle \lim_{y \to 0}\frac{\tan 2y}{y}.$$

Step 3. Make the denom­i­na­tor match the angle $2y$.

$$\displaystyle \frac{\tan 2y}{y} = 2 \cdot \frac{\tan 2y}{2y} \to 2 \cdot 1 = 2.$$

Check­ing the answer#

At $\displaystyle x = \tfrac{\pi}{2} + 0.01$: $\tan(\pi + 0.02) = \tan 0.02 \approx 0.0200$, divided by $0.01$ gives $\approx 2.00$. ✓

Answer#

$2$

Ques­tion 6: Find­ing con­stants from a limit con­di­tion#

The prob­lem#

Find $a$ and $b$ if $\displaystyle \lim_{x \to 1} f(x) = f(1)$ where $f(x) = 3x + a$ for $x < 1$, $f(1) = 5$, and $f(x) = bx - 2$ for $x > 1$.

Under­stand­ing the prob­lem#

$f$ is given by dif­fer­ent for­mu­las on the left and on the right of $x = 1$. You are told the limit at $1$ exists and equals $f(1) = 5$. That forces both one-sided lim­its to equal $5$.

The idea#

A limit exists exactly when the left-hand and right-hand lim­its exist and are equal. So set each one-sided limit equal to $5$ and solve.

Step-by-step solu­tion#

Step 1. Left-hand limit: for $x < 1$, $f(x) = 3x + a$ (a poly­no­mial, so sub­sti­tute).

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

Step 2. Right-hand limit: for $x > 1$, $f(x) = bx - 2$.

$$\displaystyle \lim_{x \to 1^+} f(x) = b(1) - 2 = b - 2.$$

Step 3. Both must equal $f(1) = 5$.

$$3 + a = 5 \Rightarrow a = 2, \qquad b - 2 = 5 \Rightarrow b = 7.$$

Check­ing the answer#

With $a = 2$, $b = 7$: left for­mula $3x + 2$ gives $5$ at $x = 1$, right for­mula $7x - 2$ gives $5$ at $x = 1$. Both sides meet at $f(1) = 5$. ✓

Answer#

$a = 2$, $b = 7$

Ques­tion 7: Deriv­a­tive of $x^3$ from first prin­ci­ples#

The prob­lem#

Find the deriv­a­tive of $f(x) = x^3$ from first prin­ci­ples and hence the slope of $y = x^3$ at $(-2, -8)$.

Under­stand­ing the prob­lem#

"From first prin­ci­ples" means you must use the def­i­n­i­tion $\displaystyle f'(x) = \lim_{h \to 0}\tfrac{f(x + h) - f(x)}{h}$, not the power rule. Then the slope of the tan­gent at a point is the value of the deriv­a­tive there.

The idea#

Expand $(x + h)^3$, sub­tract $x^3$, can­cel the com­mon fac­tor $h$, then let $h \to 0$.

Step-by-step solu­tion#

Step 1. Write the dif­fer­ence quo­tient.

$$\displaystyle \frac{f(x + h) - f(x)}{h} = \frac{(x + h)^3 - x^3}{h}.$$

Step 2. Expand $(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ and sub­tract $x^3$.

$$\displaystyle \frac{3x^2h + 3xh^2 + h^3}{h}.$$

Step 3. Can­cel $h$ (allowed since $h \ne 0$).

$$3x^2 + 3xh + h^2.$$

Step 4. Let $h \to 0$.

$$\displaystyle f'(x) = \lim_{h \to 0}(3x^2 + 3xh + h^2) = 3x^2.$$

Step 5. The point $(-2, -8)$ has $x = -2$. The slope there is

$$f'(-2) = 3(-2)^2 = 3 \times 4 = 12.$$

Check­ing the answer#

The power rule gives $\displaystyle \tfrac{d}{dx}x^3 = 3x^2$, the same. The slope is pos­i­tive, which fits: $y = x^3$ is always ris­ing.

Answer#

$f'(x) = 3x^2$; slope at $(-2, -8)$ is $12$.

Ques­tion 8: Quo­tient rule#

The prob­lem#

Dif­fer­en­ti­ate $\displaystyle \frac{x^3 - 1}{x^2 + 1}$.

Under­stand­ing the prob­lem#

The func­tion is one expres­sion divided by another, so you need the quo­tient rule.

The idea#

With $u = x^3 - 1$ and $v = x^2 + 1$, use $\displaystyle \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate the top and bot­tom.

$$u = x^3 - 1,\ u' = 3x^2; \qquad v = x^2 + 1,\ v' = 2x.$$

Step 2. Sub­sti­tute into the quo­tient rule.

$$\displaystyle \frac{dy}{dx} = \frac{3x^2(x^2 + 1) - (x^3 - 1)(2x)}{(x^2 + 1)^2}.$$

Step 3. Expand the numer­a­tor care­fully.

$$\begin{aligned} 3x^2(x^2 + 1) &= 3x^4 + 3x^2,\\ (x^3 - 1)(2x) &= 2x^4 - 2x,\\ \text{difference} &= 3x^4 + 3x^2 - 2x^4 + 2x = x^4 + 3x^2 + 2x. \end{aligned}$$

Step 4. Write the result.

$$\displaystyle \frac{dy}{dx} = \frac{x^4 + 3x^2 + 2x}{(x^2 + 1)^2}.$$

Check­ing the answer#

At $x = 0$ the for­mula gives $0$. Directly, near $x = 0$ the func­tion is $\displaystyle \tfrac{-1 + x^3}{1 + x^2} \approx -1 + (\text{terms in } x^2, x^3)$, which is flat at $0$. ✓

Answer#

$\displaystyle \frac{x^4 + 3x^2 + 2x}{(x^2 + 1)^2}$

Com­mon mis­take to avoid#

Watch the minus sign: $-(x^3 - 1)(2x) = -2x^4 + 2x$, so the $2x$ is plus.

Ques­tion 9: Prod­uct rule#

The prob­lem#

Dif­fer­en­ti­ate $(x^2 + 1)\cos x$.

Under­stand­ing the prob­lem#

This is a prod­uct of two func­tions of $x$, so use the prod­uct rule.

The idea#

With $u = x^2 + 1$ and $v = \cos x$, $(uv)' = u'v + uv'$. Remem­ber $\displaystyle \tfrac{d}{dx}\cos x = -\sin x$.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate each fac­tor.

$$u' = 2x, \qquad v' = -\sin x.$$

Step 2. Apply the prod­uct rule.

$$\displaystyle \frac{dy}{dx} = 2x\cos x + (x^2 + 1)(-\sin x) = 2x\cos x - (x^2 + 1)\sin x.$$

Check­ing the answer#

At $x = 0$: the for­mula gives $0 - 1 \cdot 0 = 0$. The func­tion $(x^2 + 1)\cos x$ is even, so its graph is sym­met­ric about the $y$-axis and must be flat at $x = 0$. ✓

Answer#

$2x\cos x - (x^2 + 1)\sin x$

Ques­tion 10: A quo­tient that sim­pli­fies nicely#

The prob­lem#

Dif­fer­en­ti­ate $\displaystyle \frac{\sin x - x\cos x}{x\sin x + \cos x}$.

Under­stand­ing the prob­lem#

It is a quo­tient, and both the top and bot­tom con­tain prod­ucts, so you will use the prod­uct rule inside the quo­tient rule. The work looks heavy, but a lot can­cels.

The idea#

Let $u = \sin x - x\cos x$ and $v = x\sin x + \cos x$. Find $u'$ and $v'$ first (each sim­pli­fies to one term), then apply the quo­tient rule and use $\sin^2 x + \cos^2 x = 1$.

Step-by-step solu­tion#

Step 1. Find $u'$. The deriv­a­tive of $x\cos x$ by the prod­uct rule is $\cos x - x\sin x$.

$$u' = \cos x - (\cos x - x\sin x) = x\sin x.$$

Step 2. Find $v'$. The deriv­a­tive of $x\sin x$ is $\sin x + x\cos x$.

$$v' = (\sin x + x\cos x) - \sin x = x\cos x.$$

Step 3. Write the quo­tient rule numer­a­tor $u'v - uv'$.

$$x\sin x\,(x\sin x + \cos x) - (\sin x - x\cos x)\,x\cos x.$$

Step 4. Expand each prod­uct.

$$\begin{aligned} x\sin x\,(x\sin x + \cos x) &= x^2\sin^2 x + x\sin x\cos x,\\ (\sin x - x\cos x)\,x\cos x &= x\sin x\cos x - x^2\cos^2 x. \end{aligned}$$

Step 5. Sub­tract. The $x\sin x\cos x$ terms can­cel.

$$x^2\sin^2 x + x\sin x\cos x - x\sin x\cos x + x^2\cos^2 x = x^2(\sin^2 x + \cos^2 x) = x^2.$$

Step 6. Divide by $v^2$.

$$\displaystyle \frac{dy}{dx} = \frac{x^2}{(x\sin x + \cos x)^2}.$$

Check­ing the answer#

At $x = 0$: the for­mula gives $\displaystyle \tfrac{0}{1} = 0$. The orig­i­nal func­tion near $0$ is $\displaystyle \tfrac{\sin x - x\cos x}{\cos x + x\sin x}$; its numer­a­tor starts like $\displaystyle \tfrac{x^3}{3}$, so the func­tion is flat at $0$. ✓

Answer#

$\displaystyle \frac{x^2}{(x\sin x + \cos x)^2}$

Ques­tion 11: Veloc­ity of a mov­ing par­ti­cle#

The prob­lem#

A par­ti­cle moves so that $s = t^3 - 6t^2 + 9t$ metres after $t$ sec­onds. Find its veloc­ity at $t = 2$ and when it is momen­tar­ily at rest.

Under­stand­ing the prob­lem#

$s$ is the posi­tion (dis­tance in metres) at time $t$. Veloc­ity is the rate of change of posi­tion, $\displaystyle v = \tfrac{ds}{dt}$. "Momen­tar­ily at rest" means the veloc­ity is zero at that instant.

The idea#

Dif­fer­en­ti­ate $s$ to get $v$. Sub­sti­tute $t = 2$; then solve $v = 0$.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate term by term with the power rule.

$$\displaystyle v = \frac{ds}{dt} = 3t^2 - 12t + 9.$$

Step 2. Veloc­ity at $t = 2$.

$$v(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3 \text{ m/s}.$$

The minus sign means the par­ti­cle is mov­ing in the neg­a­tive direc­tion (back towards the start) at that moment.

Step 3. Set $v = 0$ and divide by $3$.

$$3t^2 - 12t + 9 = 0 \quad\Longrightarrow\quad t^2 - 4t + 3 = 0.$$

Step 4. Fac­torise.

$$(t - 1)(t - 3) = 0 \quad\Longrightarrow\quad t = 1 \text{ or } t = 3.$$

Check­ing the answer#

$v(1) = 3 - 12 + 9 = 0$ ✓ and $v(3) = 27 - 36 + 9 = 0$ ✓. Also $t = 2$ lies between $1$ and $3$, where $v < 0$, match­ing $v(2) = -3$.

Answer#

Veloc­ity at $t = 2$ is $-3$ m/s; the par­ti­cle is at rest at $t = 1$ s and $t = 3$ s.

Ques­tion 12: Why $\lvert x \rvert$ has no deriv­a­tive at $0$#

The prob­lem#

For $f(x) = \lvert x \rvert$, show that $\displaystyle \lim_{h \to 0}\tfrac{f(h) - f(0)}{h}$ does not exist, so $\lvert x \rvert$ has no deriv­a­tive at $0$.

Under­stand­ing the prob­lem#

This is a "show that" ques­tion. You must prove that the limit defin­ing $f'(0)$ does not exist. A limit fails to exist when the left-hand and right-hand lim­its are dif­fer­ent.

The idea#

Sim­plify the quo­tient to $\displaystyle \tfrac{\lvert h \rvert}{h}$, then look at $h > 0$ and $h < 0$ sep­a­rately, using the def­i­n­i­tion of $\lvert h \rvert$.

Step-by-step solu­tion#

Step 1. Since $f(0) = \lvert 0 \rvert = 0$,

$$\displaystyle \frac{f(h) - f(0)}{h} = \frac{\lvert h \rvert}{h}.$$

Step 2. For $h > 0$, $\lvert h \rvert = h$, so

$$\displaystyle \frac{\lvert h \rvert}{h} = \frac{h}{h} = 1, \qquad \lim_{h \to 0^+}\frac{\lvert h \rvert}{h} = 1.$$

Step 3. For $h < 0$, $\lvert h \rvert = -h$, so

$$\displaystyle \frac{\lvert h \rvert}{h} = \frac{-h}{h} = -1, \qquad \lim_{h \to 0^-}\frac{\lvert h \rvert}{h} = -1.$$

Step 4. The right-hand limit ($1$) and the left-hand limit ($-1$) are dif­fer­ent, so the limit does not exist. There­fore $f'(0)$ does not exist.

Check­ing the answer#

Look at the graph of $y = \lvert x \rvert$: it is a V with a sharp cor­ner at the ori­gin. The right arm has slope $1$ and the left arm slope $-1$, so no sin­gle tan­gent line exists there, exactly what the two one-sided lim­its say.

Answer#

$\displaystyle \tfrac{\lvert h \rvert}{h}$ tends to $1$ from the right and $-1$ from the left, so the limit does not exist and $\lvert x \rvert$ is not dif­fer­en­tiable at $0$.