$\underset{x→0}{\lim}\frac{1}{x}(\int_y^ce^{\sin^2t}dt-\int_{x+y}^{c}e^{\sin^2t}dt)$ is equal to (where c is a constant)
Answer & explanation
Correct answer: option 1
$\underset{x→0}{\lim}\frac{1}{x}\left[\int_y^ce^{\sin^2t}dt-\int_{x+y}^{c}e^{\sin^2t}dt\right]$
From property of integration
$=\underset{x→0}{\lim}\frac{1}{x}.\left[\int_y^ce^{\sin^2t}dt+\int_{c}^{x+y}e^{\sin^2t}dt\right]$
$=\underset{x→0}{\lim}\frac{1}{x}.\int_y^{x+y}e^{\sin^2t}dt$
According to L' Hospital Rule.
$=\underset{x→0}{\lim}e^{\sin^2(x+y)}\left(1+\frac{dy}{dx}\right)-e^{\sin^2y}\frac{dy}{dx}$
$=\underset{x→0}{\lim}e^{\sin^2(x+y)}+\underset{x→0}{\lim}\frac{e^{\sin^2(x+y)}-e^{\sin^2y}}{x}.\underset{x→0}{\lim}\frac{dy}{dx}.x$
$=e^{\sin^2y}+0$
$=e^{\sin^2y}$