MDoubt No 000002 - UPSC IAS Mathematics Optional

UPSC [CSE 2019 P1]
Question 1(a) : Let $f:\left[0,\dfrac{\pi}{2}\right]\to\textbf{R}$ be a continuous function such that $$f(x)=\frac{\cos ^2 x}{4x^2-\pi^2}, 0\le x<\frac{\pi}{2}$$ Find the value of $f\left(\dfrac{\pi}{2}\right).$ 

Solution : Given that $f:\left[0,\dfrac{\pi}{2}\right]\to\textbf{R}$ is a continuous function such that
$f(x)=\dfrac{\cos ^2 x}{4x^2-\pi^2}, 0\le x<\dfrac{\pi}{2}$.
Now, since $f$ is continuous on $\left[0,\dfrac{\pi}{2}\right]$,
We have, $\displaystyle\lim_{x\to\pi/2^{-}}f(x)=f\left(\frac{\pi}{2}\right)$
$\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=\lim_{x\to\pi/2^{-}}\frac{\cos ^2 x}{4x^2-\pi^2}\quad\quad\ldots\left(\frac{0}{0}\right)$ form
$\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=\lim_{x\to\pi/2^{-}}\frac{2\cos x(-\sin x)}{8x}$
$\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=\lim_{x\to\pi/2^{-}}\frac{-2\sin 2x}{8x}$
$\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=\lim_{x\to\pi/2^{-}}\frac{-2\sin 2\left(\frac{\pi}{2}\right)}{8\left(\frac{\pi}{2}\right)}$
$\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=0$
$\therefore$ The value of $f\left(\dfrac{\pi}{2}\right)=0$.