Why do we use first principles to compute instead of just plugging into ?

Because is defined piecewise, the formula is only valid for . To find the derivative at the transition point , we must use the fundamental definition of the derivative.

JEE MainMathematics

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Limits, Continuity and Differentiability: Mathematics | JEE Main

Let

f(x)=\left\{\begin{matrix} x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0 \end{matrix}\right.

Then at

x=0

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

f

is continuous but not differentiableB

f

and

f^{\prime}

both are continuousC

f^{\prime}

is continuous but not differentiableD

f

is continuous but

f^{\prime}

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^{\prime}

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^{\prime}

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 $f'(x) = 2x\sin(1/x) - \cos(1/x)$ ; 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^{\prime}

is not continuous

None. This is the correct conclusion. Since $f'(0) = 0$ , $f$ is differentiable at $x=0$ and therefore continuous there. However, $\lim_{x \to 0} f'(x) = \lim_{x \to 0} [2x\sin(1/x) - \cos(1/x)]$ 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 $f(x) = x^2 \sin\left(\frac{1}{x}\right)$ 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 $x = 0$ : $\lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$ . Since $-1 \le \sin\left(\frac{1}{x}\right) \le 1$ , by the Squeeze Theorem, $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 = f(0)$ . Thus, $f$ is continuous at $x = 0$ .

05execute

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

06execute

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

07execute

Check continuity of $f'$ at $x = 0$ : $\lim_{x \to 0} f'(x) = \lim_{x \to 0} \left(2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)\right)$ . While $\lim_{x \to 0} 2x \sin\left(\frac{1}{x}\right) = 0$ , $\lim_{x \to 0} \cos\left(\frac{1}{x}\right)$ 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 (3).

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 $x \neq 0$ ?
$f'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)$
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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $f'(x) = 2x\sin(1/x) - \cos(1/x)$ 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 $f^{\prime}$ 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: $f^{\prime}$ 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 $f'(x) = 2x\sin(1/x) - \cos(1/x)$ ; 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 $f^{\prime}$ is not continuous - None. This is the correct conclusion. Since $f'(0) = 0$ , $f$ is differentiable at $x=0$ and therefore continuous there. However, $\lim_{x \to 0} f'(x) = \lim_{x \to 0} [2x\sin(1/x) - \cos(1/x)]$ 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 $f(x) = x^2 \sin\left(\frac{1}{x}\right)$ 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 $x = 0$ : $\lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$ . Since $-1 \le \sin\left(\frac{1}{x}\right) \le 1$ , by the Squeeze Theorem, $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 = f(0)$ . Thus, $f$ is continuous at $x = 0$ .
execute: Find $f'(0)$ using the limit definition of the derivative: $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right) = 0$ . Therefore, $f$ is differentiable at $x = 0$ , and $f'(0) = 0$ .
execute: Compute $f'(x)$ for $x \neq 0$ using the product rule: $f'(x) = \frac{d}{dx}\left[x^2 \sin\left(\frac{1}{x}\right)\right] = 2x \sin\left(\frac{1}{x}\right) + x^2 \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)$ .
execute: Check continuity of $f'$ at $x = 0$ : $\lim_{x \to 0} f'(x) = \lim_{x \to 0} \left(2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)\right)$ . While $\lim_{x \to 0} 2x \sin\left(\frac{1}{x}\right) = 0$ , $\lim_{x \to 0} \cos\left(\frac{1}{x}\right)$ 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 (3).

Limits, Continuity and Differentiability: Mathematics | JEE Main

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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