Question

${lim}_{x\to 0}\frac{1}{x}\left({\int }_y^c{e}^{{\sin }^{2}t}dt-{\int }_{x+y}^c{e}^{{\sin }^{2}t}dt\right)$ is equal to (Where $c$ is a constant)

• A. $e^{{\sin }^{2}y}$
• B. $\sin 2y{e}^{{\sin }^{2}t}$
• C. $0$
• D. None of these

Answer 1

The correct option is A $e^{{\sin }^{2}y}$

${lim}_{x\to 0}\frac{1}{x}\left({\int }_y^c{e}^{{\sin }^{2}t}dt-{\int }_{x+y}^c{e}^{{\sin }^{2}t}dt\right)={lim}_{x\to 0}\frac{{\int }_y^{x+y}{e}^{{\sin }^{2}t}dt}{x}$

According to L'hospital's Rule,

$\Rightarrow {lim}_{x\to 0}{e}^{{\sin }^2(x+y)}(1+\frac{dy}{dx})-e^{{\sin }^{2}y}\frac{dy}{dx}$

$\Rightarrow {lim}_{x\to 0}{e}^{{\sin }^2(x+y)}+\left({lim}_{x\to 0}\frac{e^{{\sin }^2(x+y)}-e^{{\sin }^{2}y}}{x}\right){lim}_{x\to 0}\frac{dy}{dx}x$

$\Rightarrow e^{{\sin }^{2}y}+0$

So, Option A is the correct answer.