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 $g\left(x\right)={\int }_x^{\frac{\pi }{2}}\left[f^{\prime }\right(t)\text{cosec}t-\cot t\text{cosec}tf(t\left)\right]dt$ for $x\in (0,\frac{\pi }{2}]$, then $\underset{x\to 0}{lim}g\left(x\right)=$

Answer 1

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

Using integration by parts for the first integral taking $u\left(t\right)=\text{cosec}t$ and $v\left(t\right)=f^{\prime }\left(t\right)$,

$g\left(x\right)=\text{cosec}tf\left(t\right){|}_x^{\frac{\pi }{2}}+{\int }_x^{\frac{\pi }{2}}\cot t\text{cosec}tf\left(t\right)dt-{\int }_x^{\frac{\pi }{2}}\cot t\text{cosec}tf\left(t\right)dt$

$g\left(x\right)=f\left(\frac{\pi }{2}\right)\text{cosec}\left(\frac{\pi }{2}\right)-f\left(x\right)\text{cosec}\left(x\right)=3-f\left(x\right)\text{cosec}x=3-\frac{f\left(x\right)}{\sin x}$

Now, $L=\underset{x\to 0}{lim}g\left(x\right)=3-\underset{x\to 0}{lim}\frac{f\left(x\right)}{\sin x}$

Using L'Hospital's rule,

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

$L=3-f^{\prime }\left(0\right)=3-1=2$