Evaluating $\lim_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x}$ 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?
 A: 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*}
A: Setting $\displaystyle\frac\pi4-x=2h$
$\displaystyle\tan x=\tan\left(\frac\pi4-2h\right)=\frac{1-\tan2h}{1+\tan2h}$
$\displaystyle\implies 1-\tan x=\frac{2\tan2h}{1+\tan2h}$ 
and $\displaystyle\sin x=\sin\left(\frac\pi4-2h\right)=\frac{\cos2h-\sin2h}{\sqrt2} $
$$\lim_{x\to\frac{\pi}4}\frac{1-\tan x}{1-\sqrt{2}\sin x} =\lim_{h\to0}\frac{2\tan2h}{(1+\tan2h)(1-\cos2h+\sin2h)} $$
$$=\lim_{h\to0}\frac{\sin2h}{1-\cos2h+\sin2h}\cdot\lim_{h\to0}\frac2{\cos2h(1+\tan2h)} $$
Now the second limit is easy, 
For the first using Double-Angle Formulas, $\displaystyle\frac{\sin2h}{1-\cos2h+\sin2h}=\frac{2\sin h\cos h}{1-(1-2\sin^2h)+2\sin h\cos h}$
$\displaystyle=\frac{\cos h}{\sin h+\cos h}$ if $\sin h\ne0$ which is true as $h\to0$
A: $$\lim_{x\to\frac{\pi}4}\frac{1-\tan x}{1-\sqrt{2}\sin x} =\frac1{\sqrt2}\lim_{x\to\frac{\pi}4}\frac{\tan\frac\pi4-\tan x}{\sin\frac\pi4-\sin x}$$
$$=\frac1{\sqrt2}\lim_{x\to\frac{\pi}4}\frac{\sin\left(\frac\pi4-x\right)}{\cos\frac\pi4\cos x}\frac1{\left(\sin\frac\pi4-\sin x\right)}$$
Now, $\displaystyle\lim_{x\to\frac{\pi}4}\frac{\sin\left(\frac\pi4-x\right)}{\frac\pi4-x}=1$
Again using Prosthaphaeresis Formula, $\displaystyle\sin\frac\pi4-\sin x=2\sin\frac{\frac\pi4-x}2\cos\frac{\frac\pi4+x}2$
Now, $\displaystyle\lim_{x\to\frac{\pi}4}\frac{\sin\left(\frac{\frac\pi4-x}2\right)}{\frac{\frac\pi4-x}2}=1$
Can you please consolidate ?
A: Another way to do it is to use Taylor expansions of $\tan(x)$ and $\sin(x)$ built around $x = \frac \pi  4$. Using only the very first terms, they write
$$\tan(x) = 1+2 \left(x-\frac{\pi }{4}\right)+2 \left(x-\frac{\pi
   }{4}\right)^2+O\left(\left(x-\frac{\pi }{4}\right)^3\right)$$
$$\sin(x) = \frac{1}{\sqrt{2}}+\frac{x-\frac{\pi }{4}}{\sqrt{2}}-\frac{\left(x-\frac{\pi
   }{4}\right)^2}{2 \sqrt{2}}+O\left(\left(x-\frac{\pi }{4}\right)^3\right)$$ Replacing in the expression
$$\frac{1-\tan x}{1-\sqrt{2}\sin x}=\frac{-2 \left(x-\frac{\pi }{4}\right)-2 \left(x-\frac{\pi
   }{4}\right)^2+O\left(\left(x-\frac{\pi }{4}\right)^3\right) } {-\left(x-\frac{\pi }{4}\right)+\frac{1}{2} \left(x-\frac{\pi
   }{4}\right)^2+O\left(\left(x-\frac{\pi }{4}\right)^3\right) }$$
Now, the limit is obvious. You can have more performing the long division and get $$\frac{1-\tan x}{1-\sqrt{2}\sin x}=2+3 \left(x-\frac{\pi }{4}\right)+O\left(\left(x-\frac{\pi }{4}\right)^2\right)$$ which shows the limit and how it is approached.
