calculus - Evaluating $lim_{xtofrac{pi}{4}}frac{1-tan x}{1-sqrt{2}sin x}$

How can I evaluate $$\lim_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x}$$ without L'Hopital rule. Using L'Hopital rule, it evaluates to 2. Is there a way to do it without using L'Hopital?

Answer

Multiply by the conjugate and use trig identities, factoring appropriately:
\begin{align*}

\lim_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x}
&= \lim_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x} \cdot \frac{1 + \sqrt{2}\sin x}{1 + \sqrt{2}\sin x} \\
&= \lim_{x\to\frac{\pi}{4}}\frac{(1-\tan x)(1 + \sqrt{2}\sin x)}{1 - 2\sin^2 x} \\
&= \lim_{x\to\frac{\pi}{4}}\frac{(1-\frac{\sin x}{\cos x})(1 + \sqrt{2}\sin x)}{(1 - \sin^2 x) - \sin^2 x} \\
&= \lim_{x\to\frac{\pi}{4}}\frac{(1-\frac{\sin x}{\cos x})(1 + \sqrt{2}\sin x)}{\cos^2 x - \sin^2 x} \cdot \frac{\cos x}{\cos x} \\
&= \lim_{x\to\frac{\pi}{4}}\frac{(\cos x - \sin x)(1 + \sqrt{2}\sin x)}{\cos x(\cos x - \sin x)(\cos x + \sin x)} \\
&= \lim_{x\to\frac{\pi}{4}}\frac{1 + \sqrt{2}\sin x}{\cos x(\cos x + \sin x)} \\
&= \frac{1 + \sqrt{2}\sin \frac{\pi}{4}}{\cos \frac{\pi}{4}(\cos \frac{\pi}{4} + \sin \frac{\pi}{4})} \\
&= \frac{1 + \sqrt{2}(\frac{1}{\sqrt 2})}{\frac{1}{\sqrt 2}(\frac{1}{\sqrt 2} + \frac{1}{\sqrt 2})}
= \frac{1 + 1}{\frac{1}{\sqrt 2}(\frac{2}{\sqrt 2})}

= \frac{2}{2/2} = 2
\end{align*}