Question

If $f\left(x\right)=\frac{\sqrt{2}-\sqrt{1+sinx}}{cosx}$ for $x\ne \pi /2$ is continuous at $x=\pi /2$ find $f\left(\pi /2\right)$

Answer 1

Since $f\left(x\right)$ is a continuous function, so limit at $x=\pi /2$ must be finite.

But at $x=\pi /2$, function is of form $\frac{0}{0}$ form, hence using L'Hopital's rule:

$f\left(\pi /2\right)=\underset{x\to \frac{\pi }{2}}{lim}\frac{\sqrt{2}-\sqrt{1+sinx}}{cosx}=$

$\underset{x\to \pi /2}{lim}\frac{-(1+sinx{)}^{\frac{-1}{2}}\times cosx}{-2sinx}=0$