Evaluate the limit: x 0 [ e^2| (x)| - 2| (x)| - 1x^2 ] — Solution | JEE Mathematics

MCQMediumJEE 2024Limits

Evaluate the limit:

$\lim_{x \to 0} \left[\frac{e^{2|\sin(x)|} - 2|\sin(x)| - 1}{x^2}\right]$

Answer

Correct answer:D

Standard Method

Given:

L = \lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{x^2}

Find: The value of the limit.

From the solution, evaluating the given limit shows that it equals $2$ .

Therefore, the correct option is D.

Expanding $e^{2|\sin x|}$ incorrectly by omitting the quadratic term is wrong, because the numerator then appears to vanish too fast. Use the expansion up to the $x^2$ -order term instead.
Ignoring the absolute value in $|\sin x|$ can be misleading, because left-hand and right-hand behavior must still be handled carefully. Near $x=0$ , use that $|\sin x|$ behaves like $|x|$ before simplifying.

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