Question

Let $f:R\to R$ be a differentiable function such that $f\left(0\right)=0,f\left(\frac{\pi }{2}\right)=3$ and $f^{\prime }\left(0\right)=1$. If $g\left(x\right)=\underset{x}{\overset{\frac{\pi }{2}}{\int }}\left[f^{\prime }\right(t)\text{cosec}t-\cot t\times \text{cosec}tf(t\left)\right]dt$ , for $x\in (0,\pi /2]$, then

$\underset{x\to 0}{lim}g\left(x\right)=$

Answer 1

$f\left(0\right)=0,f\left(\frac{\pi }{2}\right)=3,f^{\prime }\left(0\right)=1$

$g\left(x\right)={\int }_x^{\pi /2}[f^{\prime }\left(t\right)\csc t-\csc t\cot tf\left(t\right)]dt$

$={\int }_x^{\pi /2}d\left(f\left(t\right)\csc t\right)$

$\therefore g\left(x\right)=f\left(\frac{\pi }{2}\right)\csc \left(\frac{\pi }{2}\right)-f\left(x\right)\csc \left(x\right)$

$\therefore g\left(x\right)=3-f\left(x\right)\csc \left(x\right)$

$g\left(x\right)=\frac{3\sin x-f\left(x\right)}{\sin x}$

${lim}_{x\to 0}g\left(x\right)={lim}_{x\to 0}\frac{3\sin x-f\left(x\right)}{\sin x}$

As this is $\frac{0}{0}$ form, we can use L' Hospital rule

$={lim}_{x\to 0}\frac{3\cos x-f^{\prime }\left(x\right)}{\cos x}$

$=\frac{3-1}{1}$

$\therefore {lim}_{x\to 0}g\left(x\right)=2$