Question

$\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\sqrt{\cos x}-\sqrt{\sin x}}{x-{\displaystyle \frac{\mathrm{\pi }}{4}}}$

Answer 1

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\sqrt{\cos x}-\sqrt{\sin x}}{x-{\displaystyle \frac{\mathrm{\pi }}{4}}} \\ \mathrm{Rationalising}\mathrm{the}\mathrm{numerator}: \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\left(\frac{\sqrt{\cos x}-\sqrt{\sin x}}{x-{\displaystyle \frac{\mathrm{\pi }}{4}}}\right)\left(\frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\right) \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\cos x-\sin x}{\left(x-{\displaystyle \frac{\mathrm{\pi }}{4}}\right)\left(\sqrt{\cos x}+\sqrt{\sin x}\right)} \\ =\underset{h\to 0}{\lim }\frac{\cos \left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)-\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)}{\left({\displaystyle \frac{\mathrm{\pi }}{4}}+h-{\displaystyle \frac{\mathrm{\pi }}{4}}\right)\left(\sqrt{\cos \left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)}+\sqrt{\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)}\right)} \\ =\underset{h\to 0}{\lim }\frac{\cos \frac{\mathrm{\pi }}{4}\cos h-\sin \frac{\mathrm{\pi }}{4}\sin h-\sin \frac{\mathrm{\pi }}{4}\cos h-\cos \frac{\mathrm{\pi }}{4}\sin h}{h\left(\sqrt{\cos \left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)}+\sqrt{\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)}\right)} \\ =\underset{h\to 0}{\lim }\frac{-\sqrt{2}\sin h}{h\left(\sqrt{\cos \left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)}+\sqrt{\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)}\right)} \\ \Rightarrow \frac{-\sqrt{2}\times 1}{2{\left({\displaystyle \frac{1}{2}}\right)}^{{\displaystyle \frac{1}{4}}}} \\ =\frac{-1}{2^{{\displaystyle \frac{1}{4}}}}=-2^{-\frac{1}{4}} \end{gathered}$$