Question

$\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{1-\tan x}{1-\sqrt{2}\sin x}$

Answer 1

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\left[\frac{1-\tan x}{1-\sqrt{2}\sin x}\right] \\ \mathrm{It}\mathrm{is}\mathrm{of}\frac{0}{0}\mathrm{form}. \end{gathered}$$

Rationalising the denominator, we get:

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\left[\frac{\left(1-\tan x\right)\left(1+\sqrt{2}\sin x\right)}{\left(1-\sqrt{2}\sin x\right)\left(1+\sqrt{2}\sin x\right)}\right] \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\left[\frac{\left(1-\tan x\right)\left(1+\sqrt{2}\sin x\right)}{1-2{\sin }^{2}x}\right] \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\left[\frac{\left(1-{\displaystyle \frac{\sin x}{\cos x}}\right)\left(1+\sqrt{2}\sin x\right)}{\cos 2x}\right] \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\left[\frac{\left(\cos x-\sin x\right)\left(1+\sqrt{2}\sin x\right)}{\cos x\cos 2x}\right] \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\left[\frac{\left(\cos x-\sin x\right)\left(1+\sqrt{2}\sin x\right)}{\cos x·\left({\cos }^{2}x-{\sin }^{2}x\right)}\right] \\ =\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\left[\frac{\left(\cos x-\sin x\right)\left(1+\sqrt{2}\sin x\right)}{\cos x\left[\cos x-\sin x\right]\left[\cos x+\sin x\right]}\right] \\ =\frac{\left(1+\sqrt{2}\times {\displaystyle \frac{1}{\sqrt{2}}}\right)}{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)\left({\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{1}{\sqrt{2}}}\right)} \\ =\frac{2}{{\displaystyle \frac{1}{\sqrt{2}}}\times \sqrt{2}} \\ =2 \end{gathered}$$