Let $f:\mathit{R}\to \mathit{R}$ be a differentiable function such that $f\left(\frac{\pi }{4}\right)=\sqrt{2},f\left(\frac{\pi }{2}\right)=0$ and $f^{\text{'}}\left(\frac{\pi }{2}\right)=1$ and let $g\left(x\right)={\int }_x^{\frac{\pi }{4}}\left(f^{\text{'}}\left(t\right)\sec t+\tan t\sec tf\left(t\right)\right)dt$ for $x\in \left[\frac{\pi }{4},\frac{\pi }{2}\right).$ Then $\underset{x\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }g\left(x\right)$ is equal to

Given,

$g\left(x\right)={\int }_x^{\frac{\pi }{4}}\left(f^{\text{'}}\left(t\right)\sec t+\tan t\sec tf\left(t\right)\right)dt$

$\Rightarrow g\left(x\right)={\int }_x^{\frac{\pi }{4}}d\left(f\left(t\right)·\sec t\right)$$\Rightarrow g\left(x\right)={\left[f\left(t\right)\sec t\right]}_x^{\frac{\pi }{4}}$

$g\left(x\right)=f\left(\frac{\pi }{4}\right)\sec \frac{\pi }{4}-f\left(x\right)·\sec x$

$g\left(x\right)=2-f\left(x\right)\sec x=2-\left(\frac{f\left(x\right)}{\cos x}\right)$

Now taking limit both side, we get

$\underset{x\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }g\left(x\right)=2-\underset{x\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }\left(\frac{f\left(x\right)}{\cos x}\right)$

Using L'Hospital Rule

$=2-\underset{x\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }\frac{f^{\text{'}}\left(x\right)}{\left(-\sin x\right)}$

$=2+\frac{f^{\text{'}}\left(\frac{\pi }{2}\right)}{\sin \frac{\pi }{2}}=2+\frac{1}{1}=3$