JEE MainMathematics

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Differentiability and Continuity of the Derivative for Oscillatory Functions

Q: Why do we use first principles to compute f'(0) instead of just plugging x=0 into f'(x)?

Because f(x) is defined piecewise, the formula f'(x) = 2xsin(1/x) - cos(1/x) is only valid for x ≠ 0. To find the derivative at the transition point x=0, we must use the fundamental definition of the derivative.

Let f(x)={[ x^2 sin (1/x), x ≠ 0; 0, x=0 ]. Then at x=0

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Af is continuous but not differentiableBf and f' both are continuousCf' is continuous but not differentiableDf is continuous but f' is not continuous

Step-by-step solutionView

Correct answer

The function $f$ is differentiable at $x = 0$ with $f^{'} (0) = 0$ , making it continuous at $x = 0$ , but the limit $\lim_{x \to 0} f^{'} (x)$ does not exist, so $f^{'}$ is not continuous at $x = 0$ .

Option analysis

Why each option works or fails

A · f is continuous but not differentiable

Believing that the wild oscillation of $\sin (1 / x)$ prevents the difference quotient from converging to a limit as $x \to 0$ . Evaluate $f^{'} (0)$ using the definition of the derivative: $\lim_{h \to 0} \frac{h^{2} \sin (1 / h)}{h} = \lim_{h \to 0} h \sin (1 / h) = 0$ by the squeeze theorem.

B · f and f' both are continuous

Assuming that because $f^{'} (0)$ exists, $f^{'}$ must automatically be continuous at $x = 0$ , conflating differentiability of $f$ with continuity of $f^{'}$ . Differentiability at a point only guarantees the existence of the derivative value at that point; check continuity of $f^{'}$ separately by testing if $\lim_{x \to 0} f^{'} (x) = f^{'} (0)$ .

C · f' is continuous but not differentiable

Conflating the continuity of $f$ with the continuity of $f^{'}$ , mistakenly concluding $f^{'}$ is continuous. Differentiate $f (x)$ for $x \neq 0$ to get $f^{'} (x) = 2 x \sin (1 / x) - \cos (1 / x)$ ; since $\cos (1 / x)$ oscillates between $- 1$ and $1$ without settling, $\lim_{x \to 0} f^{'} (x)$ does not exist, so $f^{'}$ cannot be continuous.

D · f is continuous but f' is not continuous

This is the correct conclusion. Since $f^{'} (0) = 0$ , $f$ is differentiable at $x = 0$ and therefore continuous there. However, $\lim_{x \to 0} f^{'} (x) = \lim_{x \to 0} [2 x \sin (1 / x) - \cos (1 / x)]$ does not exist due to the $\cos (1 / x)$ term, so $f^{'}$ is discontinuous at $x = 0$ .

Reviewed route

Solution

StepWorking

01Given

Given the function $f (x) = x^{2} \sin (\frac{1}{x})$ for $x \neq 0$ and $f (0) = 0$ .

02Goal

Determine whether $f (x)$ and $f^{'} (x)$ are continuous and/or differentiable at $x = 0$ .

03Approach

Check continuity of $f$ using $\lim_{x \to 0} f (x)$ , compute $f^{'} (0)$ from the first-principles limit definition, find $f^{'} (x)$ for $x \neq 0$ via standard derivative rules, and examine whether $\lim_{x \to 0} f^{'} (x) = f^{'} (0)$ .

04Execute

Test continuity of $f$ at $x = 0$ : $\lim_{x \to 0} f (x) = \lim_{x \to 0} x^{2} \sin (\frac{1}{x})$ . Since $- 1 \leq \sin (\frac{1}{x}) \leq 1$ , by the Squeeze Theorem, $\lim_{x \to 0} x^{2} \sin (\frac{1}{x}) = 0 = f (0)$ . Thus, $f$ is continuous at $x = 0$ .

05Execute

Find $f^{'} (0)$ using the limit definition of the derivative: $f^{'} (0) = \lim_{h \to 0} \frac{f (h) - f (0)}{h} = \lim_{h \to 0} \frac{h^{2} \sin (1 / h) - 0}{h} = \lim_{h \to 0} h \sin (\frac{1}{h}) = 0$ . Therefore, $f$ is differentiable at $x = 0$ , and $f^{'} (0) = 0$ .

06Execute

Compute $f^{'} (x)$ for $x \neq 0$ using the product rule: $f^{'} (x) = \frac{d}{d x} [x^{2} \sin (\frac{1}{x})] = 2 x \sin (\frac{1}{x}) + x^{2} \cos (\frac{1}{x}) \cdot (- \frac{1}{x^{2}}) = 2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x})$ .

07Execute

Check continuity of $f^{'}$ at $x = 0$ : $\lim_{x \to 0} f^{'} (x) = \lim_{x \to 0} (2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x}))$ . While $\lim_{x \to 0} 2 x \sin (\frac{1}{x}) = 0$ , $\lim_{x \to 0} \cos (\frac{1}{x})$ does not exist (it oscillates between $- 1$ and $1$ ). Hence, $\lim_{x \to 0} f^{'} (x)$ does not exist, which means $f^{'}$ is not continuous at $x = 0$ .

✓Verify

$f$ is continuous and differentiable at $0$ , but $f^{'}$ fails to be continuous at $0$ . This matches Option D.

Hints that build this answer step by step

What is the value of $f^{'} (0)$ evaluated from the limit definition of the derivative?
$f^{'} (0) = \lim_{h \to 0} \frac{h^{2} \sin (1 / h) - 0}{h} = \lim_{h \to 0} h \sin (1 / h) = 0$
What is the formula for $f^{'} (x)$ when $x \neq 0$ ?
$f^{'} (x) = 2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x})$
Does $\lim_{x \to 0} f^{'} (x)$ exist and equal $f^{'} (0)$ ?
No, because $\cos (1 / x)$ oscillates between $- 1$ and $1$ as $x \to 0$ , so $\lim_{x \to 0} f^{'} (x)$ does not exist.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we use first principles to compute

f^{'} (0)

instead of just plugging

x = 0

into

f^{'} (x)

Because $f (x)$ is defined piecewise, the formula $f^{'} (x) = 2 x \sin (1 / x) - \cos (1 / x)$ is only valid for $x \neq 0$ . To find the derivative at the transition point $x = 0$ , we must use the fundamental definition of the derivative.

Answer

The function $f$ is differentiable at $x = 0$ with $f^{'} (0) = 0$ , making it continuous at $x = 0$ , but the limit $\lim_{x \to 0} f^{'} (x)$ does not exist, so $f^{'}$ is not continuous at $x = 0$ .

Why each option works or fails

A: f is continuous but not differentiable - Believing that the wild oscillation of $\sin (1 / x)$ prevents the difference quotient from converging to a limit as $x \to 0$ . Evaluate $f^{'} (0)$ using the definition of the derivative: $\lim_{h \to 0} \frac{h^{2} \sin (1 / h)}{h} = \lim_{h \to 0} h \sin (1 / h) = 0$ by the squeeze theorem.
B: f and f' both are continuous - Assuming that because $f^{'} (0)$ exists, $f^{'}$ must automatically be continuous at $x = 0$ , conflating differentiability of $f$ with continuity of $f^{'}$ . Differentiability at a point only guarantees the existence of the derivative value at that point; check continuity of $f^{'}$ separately by testing if $\lim_{x \to 0} f^{'} (x) = f^{'} (0)$ .
C: f' is continuous but not differentiable - Conflating the continuity of $f$ with the continuity of $f^{'}$ , mistakenly concluding $f^{'}$ is continuous. Differentiate $f (x)$ for $x \neq 0$ to get $f^{'} (x) = 2 x \sin (1 / x) - \cos (1 / x)$ ; since $\cos (1 / x)$ oscillates between $- 1$ and $1$ without settling, $\lim_{x \to 0} f^{'} (x)$ does not exist, so $f^{'}$ cannot be continuous.
D · correct: f is continuous but f' is not continuous - This is the correct conclusion. Since $f^{'} (0) = 0$ , $x = 0$ is differentiable at $x = 0$ and therefore continuous there. However, $f^{'}$ does not exist due to the $\cos (1 / x)$ term, so $f^{'}$ is discontinuous at $x = 0$ .

Step-by-step solution

Given: Given the function $f (x) = x^{2} \sin (\frac{1}{x})$ for $x \neq 0$ and $f (0) = 0$ .
Goal: Determine whether $f (x)$ and $f^{'} (x)$ are continuous and/or differentiable at $x = 0$ .
Approach: Check continuity of $f$ using $\lim_{x \to 0} f (x)$ , compute $f^{'} (0)$ from the first-principles limit definition, find $f^{'} (x)$ for $x \neq 0$ via standard derivative rules, and examine whether $\lim_{x \to 0} f^{'} (x) = f^{'} (0)$ .
Execute: Test continuity of $f$ at $x = 0$ : $\lim_{x \to 0} f (x) = \lim_{x \to 0} x^{2} \sin (\frac{1}{x})$ . Since $- 1 \leq \sin (\frac{1}{x}) \leq 1$ , by the Squeeze Theorem, $\lim_{x \to 0} x^{2} \sin (\frac{1}{x}) = 0 = f (0)$ . Thus, $f$ is continuous at $x = 0$ .
Execute: Find $f^{'} (0)$ using the limit definition of the derivative: $f^{'} (0) = \lim_{h \to 0} \frac{f (h) - f (0)}{h} = \lim_{h \to 0} \frac{h^{2} \sin (1 / h) - 0}{h} = \lim_{h \to 0} h \sin (\frac{1}{h}) = 0$ . Therefore, $f$ is differentiable at $x = 0$ , and $f^{'} (0) = 0$ .
Execute: Compute $f^{'} (x)$ for $x \neq 0$ using the product rule: $f^{'} (x) = \frac{d}{d x} [x^{2} \sin (\frac{1}{x})] = 2 x \sin (\frac{1}{x}) + x^{2} \cos (\frac{1}{x}) \cdot (- \frac{1}{x^{2}}) = 2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x})$ .
Execute: Check continuity of $f^{'}$ at $x = 0$ : $\lim_{x \to 0} f^{'} (x) = \lim_{x \to 0} (2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x}))$ . While $\lim_{x \to 0} 2 x \sin (\frac{1}{x}) = 0$ , $\lim_{x \to 0} \cos (\frac{1}{x})$ does not exist (it oscillates between $- 1$ and $1$ ). Hence, $\lim_{x \to 0} f^{'} (x)$ does not exist, which means $f^{'}$ is not continuous at $x = 0$ .
Verify: $f$ is continuous and differentiable at $0$ , but $f^{'}$ fails to be continuous at $0$ . This matches Option D.

Differentiability and Continuity of the Derivative for Oscillatory Functions

What feels right?

The function $f$ is differentiable at $x = 0$ with $f^{'} (0) = 0$ , making it continuous at $x = 0$ , but the limit $\lim_{x \to 0} f^{'} (x)$ does not exist, so $f^{'}$ is not continuous at $x = 0$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

07Execute

✓Verify

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