JEE MATHEMATICS MCQ: LIMITS

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Application +4 | -1

Let

f(x)=\left\{\begin{array}{cl}e^{-\frac{1}{x^2}} \sin \frac{1}{x} & x \neq 0, \text { then } f^{\prime}(0)= \\ 0 & x=0\end{array}\right.

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Correct option: C

Option C (0) is correct because the derivative at x=0 is found using the limit definition and the squeeze theorem.

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