Let  $f:R\to R$ be a differential 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  $g\left(x\right)=\underset{x}{\overset{\pi /4}{\int }}(f\text{'}\left(t\right)\sec \left(t\right)+\sec \left(t\right)\tan \left(t\right)f\left(t\right))dtforx\in [\frac{\pi }{4},\frac{\pi }{2})$.Then,  $\underset{x\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }g\left(x\right)$  is equal to 

$$\begin{gathered} g\left(x\right)=\underset{x}{\overset{\pi /4}{\int }}\left(f\text{'}\left(t\right)\sec t+\tan t\sec tf\left(t\right)\right)dt \\ \therefore g\left(x\right)=\underset{x}{\overset{\pi /4}{\int }}d(f\left(t\right).\sec t)={\left[f\left(t\right)\sec t\right]}_x^{\pi /4} \\ \Rightarrow g\left(x\right)=f\left(\frac{\pi }{4}\right)\sec \frac{\pi }{4}-f\left(x\right).\sec x=2-\frac{f\left(x\right)}{\cos x} \\ \therefore \underset{x\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }f\left(x\right)=2-\underset{x\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }\frac{f\left(x\right)}{\cos x} \\ =2-\underset{x\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }\frac{f^{\text{'}}\left(x\right)}{(-\sin x)} \end{gathered}$$

(Using L’ Hospital rule)

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