Question

$\underset{x\to \mathrm{\pi }/4}{\lim }\frac{\sqrt{2}\cos x-1}{\cot x-1}$ is equal to

(a) $\frac{1}{\sqrt{2}}$

(b) $\frac{1}{2}$

(c) $\frac{1}{2\sqrt{2}}$

(d) 1

Answer 1

(b) 1/2

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\sqrt{2}\cos x-1}{\cot x-1} \\ \mathrm{Rationalising}\mathrm{the}\mathrm{numerator},\mathrm{we}\mathrm{get}: \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\left(\frac{\sqrt{2}\cos x-1}{\cot x-1}\right)\times \left(\frac{\sqrt{2}\cos x+1}{\sqrt{2}\cos x+1}\right) \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\left(2{\cos }^{2}x-1\right)}{\left(\cos x-\sin x\right)}\times \frac{\sin x}{\left(\sqrt{2}\cos x+1\right)} \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\left({\cos }^{2}x-{\sin }^{2}x\right)}{\left(\cos x-\sin x\right)}\times \frac{\sin x}{\left(\sqrt{2}\cos x+1\right)} \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\left(\cos x+\sin x\right)\sin x}{\left(\sqrt{2}\cos x+1\right)} \\ =\frac{\left(\cos \frac{\mathrm{\pi }}{4}+\sin \frac{\mathrm{\pi }}{4}\right)\sin \frac{\mathrm{\pi }}{4}}{\left(\sqrt{2}\cos \frac{\mathrm{\pi }}{4}+1\right)} \\ =\frac{\left({\displaystyle \frac{1}{\sqrt{2}}}+\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}.\frac{1}{\sqrt{2}}+1} \\ =\frac{\left({\displaystyle \frac{2}{\sqrt{2}}}\right)\times \frac{1}{\sqrt{2}}}{2} \\ =\frac{1}{2} \end{gathered}$$
The correct answer is B.