Question

Evaluate : $\underset{x\to 0}{lim}\left[\frac{\stackrel{x}{\underset{0}{\int }}{\sin }^{3}tdt}{x^4}\right]$

• A. $0.25$
• B. $0.5$
• C. $.25$
• D. $0.52$

Answer 1

The correct option is A $0.25$

According to Newton Leibniz theorem

$\frac{d}{dx}{\int }_{g\left(x\right)}^{h\left(x\right)}f\left(t\right)dt=f\left(h\right(x\left)\right)h^{\prime }\left(x\right)-f\left(g\right(x\left)\right)g^{\prime }\left(x\right)$

Now, ${lim}_{x\to 0}\left[\frac{{\int }_0^x{\sin }^{3}tdt}{x^4}\right]$

is of the form $\frac{0}{0}$

So, we will have to apply L-Hospital Rule

${lim}_{x\to 0}\left[\frac{{\int }_0^x{\sin }^{3}tdt}{x^4}\right]={lim}_{x\to 0}\left[\frac{\frac{d}{dx}{\int }_0^x{\sin }^{3}tdt}{\frac{d}{dx}{x}^4}\right]={lim}_{x\to 0}\left[\frac{{\sin }^{3}x.1-{\sin }^{3}x\times 0}{4x^3}\right]={lim}_{x\to 0}\frac{{\sin }^{3}x}{4x^3}={lim}_{x\to 0}\frac{1}{4}[\frac{\sin x}{x}{]}^3=\frac{1}{4}{.1}^3=\frac{1}{4}=0.25$

Since, ${lim}_{x\to 0}\frac{\sin x}{x}=1$

Hence, Answer is $0.25$