Question

If $f\left(x\right)=\frac{1-\sin x}{(\pi -2x{)}^2},$ for $x\ne \frac{\pi }{2}$ is continuous at $x=\frac{\pi }{2},$ find $f\left(\frac{\pi }{2}\right)$

Answer 1

$LHL={lim}_{h\to 0}f\left(\right(\frac{\pi }{2})-h)={lim}_{h\to 0}\left(\frac{1-sin\left(\right(\frac{\pi }{2})-h)}{[\pi -2(\left(\frac{\pi }{2}\right)-h){]}^2}\right)={lim}_{h\to 0}\left(\frac{1-cosh}{(\pi -\pi +2h{)}^2}\right)\times \left(\frac{1+cosh}{1+cosh}\right)={lim}_{h\to 0}\left(\frac{1^2-co{s}^{2}h}{4h^2(1+cosh)}\right)={lim}_{h\to 0}\left(\frac{si{n}^{2}h}{4h^2(1+cosh)}\right)={lim}_{h\to 0}\left(\frac{si{n}^{2}h}{4h^2}\right)\times \left(\frac{1}{(1+cosh)}\right)=\left(\frac{1}{4}\right)\times 1\times \left(\right(\frac{1}{1+1}\left)\right)=\left(\frac{1}{8}\right)\therefore$

for f(x) to be continuous

$LHL=RHL=f\left(\frac{\pi }{2}\right)$

$\therefore f\left(\frac{\pi }{2}\right)=\left(\frac{1}{8}\right)$