Question

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

Answer 1

$f\left(0\right)=0,f\left(\frac{\pi }{2}\right)=3,f^{\prime }\left(0\right)=1g\left(x\right)=\underset{x}{\overset{\frac{\pi }{2}}{\int }}\left[f^{\prime }\right(t)\text{cosec}t-\cot t\text{cosec}tf(t\left)\right]dt=\underset{x}{\overset{\frac{\pi }{2}}{\int }}\frac{d}{dt}\left[f\right(t\left)cosect\right]dt=[f\left(t\right)\text{cosec}t{]}_x^{\frac{\pi }{2}}=f\left(\frac{\pi }{2}\right)\text{cosec}\left(\frac{\pi }{2}\right)-f\left(x\right)\text{cosec}x\Rightarrow g\left(x\right)=3-\frac{f\left(x\right)}{\sin x}\underset{x\to 0}{lim}g\left(x\right)=3-\underset{x\to 0}{lim}\frac{f\left(x\right)}{\sin x}\frac{0}{0}\text{form}\to \text{Apply L'Hopital's Rule}\underset{x\to 0}{lim}g\left(x\right)=3-\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{\cos x}=3-\frac{1}{1}=2$