Question

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

• A. $\frac{1}{2}$
• B. $0$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

The given limit is of the form $\frac{0}{0}$, hence using L'Hopitals' rule:-

$\underset{x\to 0}{lim}\frac{\frac{d}{dx}{\int }_0^{x^2}co{s}^{2}tdt}{\frac{d}{dx}x\sin x}$

According Leibniz integral rule-

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

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

Hence, we get:-

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

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

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

Hence, answer is option-(C).