Question

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

• A. $1$
• B. $-1$
• C. $2$
• D. ${\log }_{e}2$

Answer 1

The correct option is A $1$

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

Hence, applying L'Hopitals' Rule:-

$\underset{x\to 0}{lim}\frac{\left(\frac{d}{dx}{x}^2\right)\cos (x^2{)}^2}{\left(\frac{d}{dx}x\right)\sin x}$

According to Leibnitz integral rule-

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

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

$=\underset{x\to 0}{lim}\frac{2x\cos x^4}{x\cos x+\sin x}$

$=\underset{x\to 0}{lim}\frac{2\cos x^4}{\cos x+\frac{\sin x}{x}}$

$=\frac{2}{1+1}=1$

Hence, the answer is option-(A).