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If $ f(x) = \begin{cases} x^2 \cos\left(\frac{\pi}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases} $, then at $ x = 0 $, $ f(x) $ is

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For functions like $f(x)=x^n g(x)$ where $g(x)$ is a bounded oscillating function near $x=0$ (like $\sin(1/x)$ or $\cos(1/x)$) and $f(0)=0$: - The function is continuous at $x=0$ if $n>0$. - The function is differentiable at $x=0$ if $n>1$.

Updated On: Mar 30, 2026

Differentiable
Continuous but not differentiable
Right differentiable only
Left differentiable only

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The Correct Option is A

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If \( f(x) = \begin{cases} x^2 \cos\left(\frac{\pi}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases} \), then at \( x = 0 \), \( f(x) \) is

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The Correct Option is A

Solution and Explanation

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