Question

The value of $\underset{x\to 0}{lim}\frac{{\int }_0^x\sin t^{2}dt}{\sin x^2}$ is

• A. $1$
• B. $0$
• C. $2$
• D. none of these

Answer 1

The correct option is B $0$

The given limit is of the form $\frac{0}{0}$

Hence, using L'Hopitals' Rule:-

$\underset{x\to 0}{lim}\frac{\frac{d}{dx}{\int }_0^x\sin t^{2}dt}{\frac{d}{dx}\sin x^2}$

According to Leibnitz integral rule,

$\frac{d}{dx}{\int }_{a\left(x\right)}^{b\left(x\right)}f\left(x\right)dx$

$=f\left(b\right(x\left)\right)\frac{d}{dx}b\left(x\right)-f\left(a\right(x\left)\right)\frac{d}{dx}a\left(x\right)$

$=\underset{x\to 0}{lim}\frac{\sin x^2}{2x\cos x^2}$

$=\underset{x\to 0}{lim}\frac{1}{2}x\frac{\sin x^2}{x^2}\frac{1}{\cos x^2}$

$=0$