Continuity and Differentiability — Worked Solutions

Ques­tion 1: Con­ti­nu­ity of a poly­no­mial#

The prob­lem#

Is $f(x) = 2x^2 - 5x + 3$ con­tin­u­ous at $x = 1$? Every­where?

Under­stand­ing the prob­lem#

$f$ is a poly­no­mial, defined for every real $x$. You must check con­ti­nu­ity at the par­tic­u­lar point $1$, and then decide about every point.

The idea#

Check the def­i­n­i­tion at $x = 1$: the limit must equal the value. Then use the lesson's fact that every poly­no­mial is con­tin­u­ous on $\mathbf{R}$.

Step-by-step solu­tion#

Step 1. Value at $1$.

$$f(1) = 2(1)^2 - 5(1) + 3 = 2 - 5 + 3 = 0.$$

Step 2. Limit at $1$. A poly­no­mi­al's limit is found by sub­sti­tu­tion, and the same expres­sion applies on both sides of $1$.

$$\displaystyle \lim_{x \to 1} f(x) = 2 - 5 + 3 = 0.$$

Step 3. The left limit, right limit and value are all $0$, so $f$ is con­tin­u­ous at $1$.

Step 4. The same argu­ment works at any real $c$, because $f$ is a poly­no­mial. So $f$ is con­tin­u­ous every­where.

Check­ing the answer#

$f(x) = (2x - 3)(x - 1)$, a smooth parabola with no breaks; its graph crosses the $x$-axis at $x = 1$, match­ing $f(1) = 0$.

Answer#

Yes, $f$ is con­tin­u­ous at $x = 1$, and it is con­tin­u­ous every­where (it is a poly­no­mial).

Ques­tion 2: A piece­wise func­tion at the join­ing point#

The prob­lem#

Exam­ine con­ti­nu­ity of $f(x) = \begin{cases} x + 4, & x < 1 \\ 5x, & x \ge 1 \end{cases}$ at $x = 1$.

Under­stand­ing the prob­lem#

The for­mula changes at $x = 1$. To the left, use $x + 4$; at $1$ and to the right, use $5x$.

The idea#

Com­pute the left-hand limit, the right-hand limit and $f(1)$, and see whether all three agree.

Step-by-step solu­tion#

Step 1. Left-hand limit (use $x + 4$).

$$\displaystyle \lim_{x \to 1^-} (x + 4) = 1 + 4 = 5.$$

Step 2. Right-hand limit (use $5x$).

$$\displaystyle \lim_{x \to 1^+} 5x = 5.$$

Step 3. Value at $1$ (the rule $x \ge 1$ applies).

$$f(1) = 5(1) = 5.$$

Step 4. All three equal $5$, so $f$ is con­tin­u­ous at $x = 1$.

Check­ing the answer#

The two straight lines $y = x + 4$ and $y = 5x$ both pass through $(1, 5)$, so the graph joins with­out a gap.

Answer#

$f$ is con­tin­u­ous at $x = 1$ (left limit $=$ right limit $= f(1) = 5$).

Ques­tion 3: Points of dis­con­ti­nu­ity#

The prob­lem#

Find all points of dis­con­ti­nu­ity of (a) $\displaystyle f(x) = \frac{x + 2}{x^2 - 9}$ and of (b) $[2x]$ on $[0, 2]$.

Under­stand­ing the prob­lem#

(a) is a ratio­nal func­tion. (b) uses the great­est inte­ger func­tion: $[t]$ is the largest inte­ger not exceed­ing $t$. It jumps at every inte­ger value of its input.

The idea#

(a) A ratio­nal func­tion is con­tin­u­ous wher­ever its denom­i­na­tor is non-zero, and it can­not be con­tin­u­ous where it is not defined. (b) $[2x]$ jumps exactly where $2x$ is an inte­ger.

Step-by-step solu­tion#

Part (a)

Step 1. Find where the denom­i­na­tor is zero.

$$x^2 - 9 = 0 \;\Rightarrow\; (x - 3)(x + 3) = 0 \;\Rightarrow\; x = 3 \text{ or } x = -3.$$

Step 2. The numer­a­tor $x + 2$ is not zero at either point, so noth­ing can­cels; $f$ is unde­fined there. Every­where else $f$ is con­tin­u­ous.

Part (b)

Step 1. $2x$ is an inte­ger when $\displaystyle x = 0, \tfrac12, 1, \tfrac32, 2$ in $[0, 2]$.

Step 2. At $\displaystyle x = \tfrac12$: for $x$ just less than $\displaystyle \tfrac12$, $2x$ is just less than $1$, so $[2x] = 0$; at and just after $\displaystyle \tfrac12$, $[2x] = 1$.

$$\displaystyle \lim_{x \to \frac12^-}[2x] = 0 \ne 1 = [2 \cdot \tfrac12].$$

So there is a jump. The same hap­pens at $x = 1$ (jump from $1$ to $2$), at $\displaystyle x = \tfrac32$ (from $2$ to $3$) and at $x = 2$ (from $3$ to $4$).

Step 3. At $x = 0$, the left end of the inter­val, only the right-hand limit mat­ters: $[2x] = 0$ for $\displaystyle 0 \le x < \tfrac12$, so the right limit is $0 = [0]$. It is con­tin­u­ous there.

Step 4. Between these points $[2x]$ is con­stant, hence con­tin­u­ous.

Check­ing the answer#

(b) The graph of $[2x]$ on $[0, 2]$ is a stair­case with steps of width $\displaystyle \tfrac12$: $0, 1, 2, 3$, and then the sin­gle value $4$ at $x = 2$. It breaks at $\displaystyle \tfrac12, 1, \tfrac32, 2$.

Answer#

(a) $x = 3$ and $x = -3$. (b) $\displaystyle x = \tfrac12, 1, \tfrac32, 2$ (it is con­tin­u­ous at $0$ from the right).

Ques­tion 4: Choos­ing $k$ to remove a gap#

The prob­lem#

Find $k$: $\displaystyle f(x) = \begin{cases} \frac{\sin 3x}{x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ con­tin­u­ous at $0$.

Under­stand­ing the prob­lem#

For $x \ne 0$ the for­mula is known; the value at $0$ is the unknown $k$. Con­ti­nu­ity at $0$ requires $\displaystyle \lim_{x \to 0} f(x) = f(0) = k$.

The idea#

Use the stan­dard limit $\displaystyle \lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1$ with $\theta = 3x$.

Step-by-step solu­tion#

Step 1. Rewrite to match the stan­dard form by mul­ti­ply­ing and divid­ing by $3$.

$$\displaystyle \frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x}.$$

Step 2. As $x \to 0$, $3x \to 0$ too.

$$\displaystyle \lim_{x \to 0} \frac{\sin 3x}{x} = 3 \cdot \lim_{3x \to 0}\frac{\sin 3x}{3x} = 3 \times 1 = 3.$$

Step 3. Set the limit equal to $f(0)$.

$$k = 3.$$

Check­ing the answer#

$x = 0.01$: $\sin(0.03) \approx 0.029996$, and $\displaystyle \frac{0.029996}{0.01} \approx 2.9996$, close to $3$.

Answer#

$k = 3$.

Ques­tion 5: Two con­stants for con­ti­nu­ity every­where#

The prob­lem#

Find $a$ and $b$: $f(x) = \begin{cases} 2, & x \le 0 \\ ax + b, & 0 < x < 4 \\ 10, & x \ge 4 \end{cases}$ con­tin­u­ous every­where.

Under­stand­ing the prob­lem#

Each piece is a poly­no­mial, so $f$ is auto­mat­i­cally con­tin­u­ous inside each inter­val. The only risk is at the join­ing points $x = 0$ and $x = 4$. Two con­di­tions give two equa­tions for $a$ and $b$.

The idea#

At each join­ing point, make the left limit, right limit and value equal.

Step-by-step solu­tion#

Step 1. At $x = 0$. Left limit and $f(0)$ use the con­stant $2$; the right limit uses $ax + b$.

$$\displaystyle \lim_{x \to 0^+}(ax + b) = b, \qquad f(0) = 2 \;\Rightarrow\; b = 2.$$

Step 2. At $x = 4$. The left limit uses $ax + b$; the right limit and $f(4)$ use $10$.

$$\displaystyle \lim_{x \to 4^-}(ax + b) = 4a + b, \qquad f(4) = 10 \;\Rightarrow\; 4a + b = 10.$$

Step 3. Sub­sti­tute $b = 2$.

$$4a + 2 = 10 \;\Rightarrow\; 4a = 8 \;\Rightarrow\; a = 2.$$

Check­ing the answer#

The mid­dle piece is $2x + 2$: at $x = 0$ it gives $2$ and at $x = 4$ it gives $10$, match­ing both outer pieces.

Answer#

$a = 2$, $b = 2$.

Ques­tion 6: $\displaystyle x^2\sin\frac1x$ at the ori­gin#

The prob­lem#

Is $\displaystyle f(x) = \begin{cases} x^2 \sin\frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$ con­tin­u­ous at $0$?

Under­stand­ing the prob­lem#

$\displaystyle \sin\frac1x$ oscil­lates wildly near $0$ and has no limit. But it is mul­ti­plied by $x^2$, which shrinks to $0$. You must decide whether $\displaystyle \lim_{x \to 0} f(x) = 0 = f(0)$.

The idea#

Use the bound $\displaystyle \left\lvert \sin\frac1x \right\rvert \le 1$ to trap $f(x)$ between $-x^2$ and $x^2$, both of which tend to $0$.

Step-by-step solu­tion#

Step 1. For $x \ne 0$, since $\lvert \sin\theta \rvert \le 1$ for every $\theta$,

$$\displaystyle \left\lvert x^2\sin\frac1x \right\rvert = x^2\left\lvert \sin\frac1x \right\rvert \le x^2.$$

Step 2. So $-x^2 \le f(x) \le x^2$. As $x \to 0$, both $-x^2 \to 0$ and $x^2 \to 0$.

Step 3. There­fore $f(x)$, trapped between them, also tends to $0$.

$$\displaystyle \lim_{x \to 0} f(x) = 0.$$

Step 4. This equals $f(0) = 0$, so $f$ is con­tin­u­ous at $0$.

Check­ing the answer#

At $x = 0.01$, $\lvert f(x) \rvert \le 0.0001$, how­ever $\sin(100)$ behaves. The val­ues really do shrink to $0$.

Answer#

Yes. Since $\displaystyle \lvert x^2\sin\frac1x \rvert \le x^2 \to 0$, the limit is $0 = f(0)$.

Ques­tion 7: $\lvert x + 2 \rvert$ at $x = -2$#

The prob­lem#

Show that $f(x) = \lvert x + 2 \rvert$ is con­tin­u­ous but not dif­fer­en­tiable at $x = -2$.

Under­stand­ing the prob­lem#

Two things must be shown: (i) con­ti­nu­ity at $-2$; (ii) the deriv­a­tive does not exist there. Write the func­tion with­out the mod­u­lus:

$$f(x) = \begin{cases} -(x + 2), & x < -2 \\ x + 2, & x \ge -2. \end{cases}$$

The idea#

For con­ti­nu­ity, com­pare the one-sided lim­its with $f(-2)$. For dif­fer­en­tia­bil­ity, com­pute the left and right dif­fer­ence quo­tients $\displaystyle \frac{f(-2 + h) - f(-2)}{h}$.

Step-by-step solu­tion#

Step 1. Value: $f(-2) = \lvert 0 \rvert = 0$.

Step 2. One-sided lim­its.

$$\displaystyle \lim_{x \to -2^-} -(x + 2) = 0, \qquad \lim_{x \to -2^+}(x + 2) = 0.$$

Both equal $f(-2)$, so $f$ is con­tin­u­ous at $-2$.

Step 3. Dif­fer­ence quo­tient: $f(-2 + h) - f(-2) = \lvert h \rvert$, so the quo­tient is $\displaystyle \frac{\lvert h \rvert}{h}$.

Step 4. Right side, $h > 0$: $\lvert h \rvert = h$.

$$\displaystyle \lim_{h \to 0^+}\frac{h}{h} = 1.$$

Step 5. Left side, $h < 0$: $\lvert h \rvert = -h$.

$$\displaystyle \lim_{h \to 0^-}\frac{-h}{h} = -1.$$

Step 6. The one-sided deriv­a­tives $-1$ and $1$ dif­fer, so $f'(-2)$ does not exist. $\blacksquare$

Check­ing the answer#

The graph is a V with its cor­ner at $(-2, 0)$: slope $-1$ on the left, slope $+1$ on the right. No sin­gle tan­gent line exists at a cor­ner.

Answer#

$f$ is con­tin­u­ous at $-2$ (both lim­its $= 0 = f(-2)$) but not dif­fer­en­tiable there (left deriv­a­tive $-1$, right deriv­a­tive $1$).

Ques­tion 8: $\lvert x \rvert + \lvert x - 1 \rvert$ at $0$ and $1$#

The prob­lem#

Is $f(x) = \lvert x \rvert + \lvert x - 1 \rvert$ dif­fer­en­tiable at $0$ and $1$?

Under­stand­ing the prob­lem#

Each mod­u­lus term has a cor­ner where its inside is $0$: $\lvert x \rvert$ at $0$, $\lvert x - 1 \rvert$ at $1$. Check each point.

The idea#

Remove the mod­u­lus signs on each inter­val to get sim­ple lin­ear pieces, then com­pare the slopes on either side of each point.

Step-by-step solu­tion#

Step 1. Write $f$ piece by piece.

$$f(x) = \begin{cases} -x - (x - 1) = 1 - 2x, & x < 0 \\ x - (x - 1) = 1, & 0 \le x < 1 \\ x + (x - 1) = 2x - 1, & x \ge 1 \end{cases}$$

Step 2. At $x = 0$: the slope just to the left is $-2$ and just to the right is $0$.

$$Lf'(0) = -2, \qquad Rf'(0) = 0.$$

These dif­fer, so $f$ is not dif­fer­en­tiable at $0$.

Step 3. At $x = 1$: the slope just to the left is $0$ and just to the right is $2$.

$$Lf'(1) = 0, \qquad Rf'(1) = 2.$$

These dif­fer, so $f$ is not dif­fer­en­tiable at $1$.

Check­ing the answer#

$f$ is con­tin­u­ous (at $0$: $1 - 0 = 1$; at $1$: $2 - 1 = 1$), so the graph is unbro­ken but has two cor­ners: it falls, runs flat at height $1$ between $0$ and $1$, then rises.

Answer#

$f$ is not dif­fer­en­tiable at $x = 0$ (slopes $-2$ and $0$) or at $x = 1$ (slopes $0$ and $2$).

Ques­tion 9: Con­tin­u­ous on $\mathbf{R}$ but not dif­fer­en­tiable at $5$#

The prob­lem#

Give a func­tion that is con­tin­u­ous on $\mathbf{R}$ but not dif­fer­en­tiable at $x = 5$.

Under­stand­ing the prob­lem#

You need an exam­ple with an unbro­ken graph that has a cor­ner at $x = 5$.

The idea#

$\lvert x \rvert$ is con­tin­u­ous every­where with a cor­ner at $0$. Shift it $5$ units to the right.

Step-by-step solu­tion#

Step 1. Take $f(x) = \lvert x - 5 \rvert$. It is a com­pos­ite of the con­tin­u­ous func­tions $x - 5$ and $\lvert \cdot \rvert$, so it is con­tin­u­ous on $\mathbf{R}$.

Step 2. At $x = 5$ the dif­fer­ence quo­tient is

$$\displaystyle \frac{f(5 + h) - f(5)}{h} = \frac{\lvert h \rvert}{h},$$

which tends to $1$ from the right and $-1$ from the left.

Step 3. The one-sided deriv­a­tives dif­fer, so $f$ is not dif­fer­en­tiable at $5$.

Check­ing the answer#

This is the same com­pu­ta­tion as in Ques­tion 7, moved to the point $5$.

Answer#

$f(x) = \lvert x - 5 \rvert$ (one exam­ple; any func­tion with an unbro­ken cor­ner at $5$ works).

Ques­tion 10: The great­est inte­ger func­tion at $2.5$ and at $2$#

The prob­lem#

Is $f(x) = [x]$ dif­fer­en­tiable at $x = 2.5$? At $x = 2$?

Under­stand­ing the prob­lem#

$[x]$ is con­stant between con­sec­u­tive inte­gers and jumps at each inte­ger. Dif­fer­en­tia­bil­ity at a point needs, first of all, con­ti­nu­ity there.

The idea#

Near $2.5$, $[x] = 2$ is con­stant, so its slope is $0$. At $2$, check con­ti­nu­ity; if it fails, dif­fer­en­tia­bil­ity fails too, since every dif­fer­en­tiable func­tion is con­tin­u­ous.

Step-by-step solu­tion#

Step 1. At $x = 2.5$: for all $x$ in $(2, 3)$, $[x] = 2$. For small $h$ (with $\lvert h \rvert < 0.5$),

$$\displaystyle \frac{f(2.5 + h) - f(2.5)}{h} = \frac{2 - 2}{h} = 0 \to 0.$$

So $f'(2.5) = 0$ and $f$ is dif­fer­en­tiable there.

Step 2. At $x = 2$: the left-hand limit is

$$\displaystyle \lim_{x \to 2^-}[x] = 1,$$

but $f(2) = 2$. The limit does not equal the value, so $f$ is not con­tin­u­ous at $2$.

Step 3. A func­tion that is not con­tin­u­ous at a point can­not be dif­fer­en­tiable there.

Check­ing the answer#

Directly, the left quo­tient $\displaystyle \frac{[2 + h] - 2}{h} = \frac{1 - 2}{h} = -\frac1h$ for small neg­a­tive $h$, which grows with­out bound. No deriv­a­tive exists.

Answer#

At $x = 2.5$: yes, $f'(2.5) = 0$. At $x = 2$: no, because $[x]$ is not even con­tin­u­ous there.