Question

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

Answer 1

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