$\mathop {\lim }\limits_{x \to 0} \frac{1}{x}\left\{ {\mathop \smallint \limits_y^a {e^{{{\sin }^2}t}}dt - \mathop \smallint \limits_{x + y}^a {e^{{{\sin }^2}t}}dt} \right\}$ is equal to (where $a$ is a constant)

$\sin \,2y\,\,{e^{{{\sin }^2}y}}$

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Analyzing +4 | -1

\lim\limits_{x \to 0} \frac{1}{x}\left\{ {\int \limits_y^a {e^{{{\sin }^2}t}}dt - \int \limits_{x + y}^a {e^{{{\sin }^2}t}}dt} \right\}

is equal to (where

a

is a constant)

View answer and explanation

Correct option: A

{e^{{{\sin }^2}y}}

Option A (

{e^{{{\sin }^2}y}}

) is correct because it is the result of applying the Leibniz rule for differentiating under the integral sign and then evaluating the limit.