Continuity and differentiability of $x^2\sin(A/x)$ ¶

Math I: Calculus · Weekly test 2025-26, Week 3, open question · 7 points · Lectures 5–6

Problem¶

Let $A,B\in\mathbb{R}$ and define $f\colon\mathbb{R}\to\mathbb{R}$ by

f(x)= \begin{cases} x^2\sin(A/x), & x\neq0,\\ B, & x=0. \end{cases}

(a) Find all $A,B$ for which $f$ is continuous at $0\text{.}$ [2 points]
(b) Find all $A,B$ for which $f$ is differentiable at $0\text{,}$ and determine $f'(0)\text{.}$ [3 points]
(c) Find all $A,B$ for which $f'$ is continuous at $0\text{.}$ [2 points]

What you need¶

Continuity: $\lim_{x\to0}f(x)=f(0)\text{.}$
Squeeze theorem for functions, with $|\sin t|\leq1\text{.}$
Derivative and differentiability: the difference quotient at $0\text{,}$ and the fact that differentiability implies continuity.
Chain rule and differentiation rules for $x\neq0\text{.}$
Limit at a point: if two sequences $x_k\to0$ and $y_k\to0$ give different limits of $f'\text{,}$ then $\lim_{x\to0}f'(x)$ does not exist.

Plan¶

The factor $x^2$ beats the oscillation of $\sin(A/x)$ in (a) and (b), so $A$ never matters there; only $B$ does. In (c) the derivative for $x\neq0$ has a term $-A\cos(A/x)$ with no $x$ in front, which keeps oscillating. Two sequences that pick out $\cos=1$ and $\cos=-1$ show the limit fails unless $A=0\text{.}$

Step-by-step solution¶

Try each step yourself before you open it.

(a) · Continuity at 0 (2 points)

For $x\neq0\text{,}$ $0\leq|x^2\sin(A/x)|\leq x^2\text{,}$ and $x^2\to0\text{.}$ By the squeeze theorem,

\lim_{x\to0}f(x)=0

for every $A\text{.}$ So $f$ is continuous at $0$ exactly when $f(0)=B=0\text{,}$ with $A$ arbitrary.

Concept: Squeeze theorem for functions

(b) Step 1 · Differentiability forces B = 0 (1 point)

A function that is differentiable at a point is continuous there. By (a), $B=0$ is necessary.

(b) Step 2 · Compute the difference quotient (2 points)

With $B=0\text{,}$

\frac{f(x)-f(0)}{x-0}=x\sin\left(\frac{A}{x}\right),

and $|x\sin(A/x)|\leq|x|\to0\text{.}$ So $f'(0)=0$ for every $A\text{.}$ Hence $f$ is differentiable at $0$ exactly when $B=0\text{,}$ and then $f'(0)=0\text{.}$

(c) Step 1 · The derivative away from 0

For $x\neq0\text{,}$ the product rule and chain rule give

f'(x)=2x\sin\left(\frac{A}{x}\right) +x^2\cos\left(\frac{A}{x}\right)\cdot\left(-\frac{A}{x^2}\right) =2x\sin\left(\frac{A}{x}\right)-A\cos\left(\frac{A}{x}\right).

The first term tends to $0\text{.}$ The second term has no factor $x$ to damp it.

(c) Step 2 · If A ≠ 0 the limit of f′ does not exist (with Step 3: 2 points)

Take, for positive integers $k\text{,}$

x_k=\frac{A}{2\pi k},\quad y_k=\frac{A}{(2k+1)\pi}.

Both tend to $0\text{.}$ At these points $\sin(A/x)=0\text{,}$ while $\cos(A/x_k)=1$ and $\cos(A/y_k)=-1\text{.}$ So

f'(x_k)=-A,\quad f'(y_k)=A.

Since $A\neq-A$ when $A\neq0\text{,}$ $\lim_{x\to0}f'(x)$ does not exist, and $f'$ is not continuous at $0\text{.}$

(c) Step 3 · The remaining case

For $f'$ to be continuous at $0$ it must first exist there, so $B=0$ by (b). If also $A=0\text{,}$ then $f\equiv0\text{,}$ so $f'\equiv0\text{,}$ which is continuous. Therefore $f'$ is continuous at $0$ exactly when $A=B=0\text{.}$

Where marks are lost¶

Using the derivative formula at $x=0\text{.}$ The formula from (c) Step 1 is only valid for $x\neq0\text{.}$ At $0$ you must use the difference quotient.
Concluding from " $\cos(A/x)$ has no limit" without proof. Give two sequences with different limits of $f'\text{,}$ as above.
Forgetting $B$ in (b) and (c). Differentiability and continuity of $f'$ both need $f$ to be continuous at $0$ first.

Try next¶

Continuity of $1/x^2$ from the $\varepsilon$ – $\delta$ definition
The inverse of $x+x^3$ : derivatives of an inverse function.

Symbols: delta, element of, epsilon, pi, real numbers

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