JEE MATHEMATICS MCQ: LIMITS

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Application +4 | -1

\lim \limits_{x \rightarrow \frac{-1}{\sqrt{2}}} \frac{f(x)-2}{2 x^2-1}=3 \pi

, then

\lim \limits_{x \rightarrow \frac{-1}{\sqrt{2}}} e^{f(x)}

will be

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the limit of f(x) as x approaches -1/sqrt(2) is 2, so the limit of e^(f(x)) is e².

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