JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

Let

P=\lim \limits_{x \rightarrow 0^{+}}\left(1+\tan ^2 \sqrt{x}\right)^{\frac{1}{2 x}}

. Then,

\log _e P

, is equal to

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

\frac{{1}}{{2}}

) is correct because the limit can be evaluated using the exponential form of limits and the standard limit

\lim_{x \rightarrow 0} \frac{{\tan x}}{{x}} = 1

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