Question

Evaluate: $\underset{x\to 0}{lim}\left[\frac{{\int }_0^x{\sin }^{3}tdt}{x^4}\right]$

• A. $0.25$
• B. $2.5$
• C. $5.2$
• D. $0.52$

Answer 1

The correct option is A $0.25$

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

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

Hence,using L'Hopitals' Rule,we get-

$\underset{x\to 0}{lim}\frac{\frac{d}{dx}{\int }_0^x{\sin }^{3}tdt}{\frac{d}{dx}{x}^4}$

Now, according Leibniz integral rule-

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

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

Applying this to the given condition-

$\underset{x\to 0}{lim}\frac{{\sin }^{3}x}{4x^3}$

$=\frac{1}{4}\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^3$

$=\frac{1}{4}=0.25$ $[\underset{x\to 0}{lim}\left(\frac{\sin x}{x}\right)=1]$.