Evaluate the following limit: _ x 4 1- 1-2

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\tan\text{x}}{1-\sqrt{2}\sin\text{x}}$

✓

Answer

$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\tan\text{x}}{1-\sqrt{2}\sin\text{x}}$
$\text{x}\rightarrow\frac{\pi}{4},$ then $\text{x}-\frac{\pi}{4}\rightarrow0,$ let $\text{x}-\frac\pi4=\text{y}$
$\Rightarrow\lim\limits_{{\text{x}\rightarrow{\frac{\pi}{4}}}}\frac{1-\tan\text{x}}{1-\sqrt{2}\sin\text{x}}=\lim\limits_{{\text{x}-{\frac{\pi}{4}}}\rightarrow0}\frac{1-\tan\text{x}}{1-\sqrt{2}\sin\text{x}}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{1-\tan\big(\text{y}+\frac{\pi}{4}\big)}{1-\sqrt{2}\sin\big(\text{y}+\frac\pi4\big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{1-\Bigg(\frac{\tan\frac\pi4+\tan\text{y}}{1+\tan\frac\pi4+\tan\text{y}}\Bigg)}{1-\sqrt{2}\Big(\sin\text{y}\cos\frac{\pi}{4}+\cos\text{y}\sin\frac\pi4\Big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\Big(1-\Big(\frac{1+\tan\text{y}}{1-\tan\text{y}}\Big)\Big)}{1-\sqrt{2}\Big(\frac{\sin\text{y}}{\sqrt{2}}+\frac{\cos\text{y}}{\sqrt{2}}\Big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{(1-\tan\text{y}-1-\tan\text{y})}{(1-\tan\text{y})(1-\sin\text{y}-\cos\text{y})}$
$=\lim\limits_{\text{y}\rightarrow0}\Big(\frac{-2\tan\text{y}}{(1-\tan\text{y})(1-\sin\text{y}-\cos\text{y})}\Big)$
$=-2\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}\times1}{\lim\limits_{\text{y}\rightarrow0}(1-\tan\text{y})\times\lim\limits_{\text{y}\rightarrow0}(1\sin\text{y}-\cos\text{y})}$
$=\frac{-2\Big(\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}}{\text{y}}\Big)\times\text{y}}{\Big(\lim\limits_{\text{y}\rightarrow0}(1)-\lim\limits_{\text{y}\rightarrow0}\tan\text{y}\Big)\times\Big(1-\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{\text{y}}\times\text{y}-\cos0\Big)}$
$=\frac{-2}{(1-\text{y})(1-\text{y}-1)}=\frac{-2\text{y}}{(1-\text{y})(-\text{y})}=\frac{2}{1-\text{y}}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{2}{1-\text{y}}=2$
$=2$