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 $1$

$\underset{x\to 0}{lim}\frac{{\int }_0^{x^2}\cos \left(t^2\right)dt}{x\sin x}$ $\left[\frac{0}{0}form\right]$

$=\underset{x\to 0}{lim}\frac{\cos \left(x^4\right)\times 2x}{\sin x+x\cos x}$ [L'Hospital's rule]

$=\underset{x\to 0}{lim}\frac{2[\cos x^4-x\sin \left(x^4\right)\times 4x^3]}{\cos x+\cos x-x\sin x}$ [L'Hospital's rule]

$=\frac{2[\cos 0-0]}{\cos 0+\cos 0-0}=\frac{2}{1+1}=1$