Question

$\stackrel{Lim}{x\to 0}\frac{{\int }_0^{x}sin{t}^{2}dt}{x(1-cosx)}$ equals

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

Answer 1

The correct option is D $\frac{2}{3}$

$\underset{x\to 0}{lim}$ $\frac{{\int }_0^x\sin t^{2}dt}{x(1-cosx)}$

Putting the limit we see that the above is in an indeterminate form $÷$ so using $L^1$ Hospital's rule

$=\underset{x\to 0}{lim}$ $\frac{\frac{d}{dx}\left({\int }_0^x\sin t^{2}dt\right)}{\frac{d}{dx}(x-xcosx)}$

Now, in the numerator, we use Leibnitz rule which states that

$\frac{d}{dx}{\int }_{a\left(x\right)}^{b\left(x\right)}f\left(x\right)dx=f\left(b\right).\frac{db}{dx}-f\left(a\right)\frac{da}{dx}$ ; Using this we can write

$=\underset{x\to 0}{lim}$ $\frac{{sinx}^2\left(1\right)-\sin \left(0^2\right).\left(0\right)}{1-(cosx-xsinx)}$

$=\underset{x\to 0}{lim}$ $\frac{\sin x^2}{1-cosx+xsinx}$

Again we can see that it is in $÷$ form, so employing $L^{{}^{\prime }}$ Hospital's rule.

$=\underset{x\to 0}{lim}$ $\frac{\cos x^2\left(2x\right)}{+sinx+sinx+xcosx}$

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

It is still in $(÷)$ form, so again employing $L^{{}^{\prime }}$ Hospital's rule

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

$=\underset{x\to 0}{lim}$ $\frac{2\cos x^2-4x^2\sin x^2}{3cosx-xsinx}$

$=\frac{2\cos \left(0\right)-0}{3\cos \left(0\right)-0}$

$=\frac{2}{3}$