In looking at the corresponding graph and differentiating it after reducing it to a different form, I know the that limit is equal to $2$ but I am unsure as to how I can show this algebraically. Any hints would be appreciated.
$$\lim_{x \rightarrow \pi/4} \frac{1-\tan x }{1-\sqrt2 \, \sin x}$$
$$\lim_{x\to\dfrac\pi4}\frac{1-\tan x}{1-\sqrt2\sin x}$$
$$=\lim_{x\to\dfrac\pi4}\frac{\cos x-\sin x}{1-\sqrt2\sin x}\cdot\frac1{\lim_{x\to\dfrac\pi4}(\cos x)}$$
$$=\lim_{x\to\dfrac\pi4}\frac{(\cos x-\sin x)(\cos x+\sin x)}{(1-\sqrt2\sin x)(1+\sqrt2\sin x)}\cdot\lim_{x\to\dfrac\pi4}\frac{(1+\sqrt2\sin x)}{(\cos x+\sin x)}\cdot\frac1{\dfrac1{\sqrt2}}$$
$$=\lim_{x\to\dfrac\pi4}\frac{\cos2x}{\cos2x}\cdot\lim_{x\to\dfrac\pi4}\frac{(1+\sqrt2\sin x)}{(\cos x+\sin x)}\cdot\sqrt2$$
as $\displaystyle(\cos x-\sin x)(\cos x+\sin x)=\cos^2x-\sin^2x=\cos2x$
and $\displaystyle(1-\sqrt2\sin x)(1+\sqrt2\sin x)=1-2\sin^2x=\cos2x$
and as $\displaystyle x\to\dfrac\pi4,x\ne\dfrac\pi4,\cos2x\ne0$
$$\lim_{x\to\dfrac\pi4}\frac{1-\tan x}{1-\sqrt2\sin x}$$
$$=\frac1{\sqrt2}\lim_{x\to\dfrac\pi4}\frac{\tan x-1}{\sin x-\dfrac1{\sqrt2}}$$
$$=\frac1{\sqrt2}\frac{\lim_{x\to\dfrac\pi4}\left(\dfrac{\tan x-\tan\dfrac\pi4}{x-\dfrac\pi4}\right)}{\lim_{x\to\dfrac\pi4}\left(\dfrac{\sin x-\sin\dfrac\pi4}{x-\dfrac\pi4}\right)}$$
$$=\frac1{\sqrt2}\cdot\frac{\dfrac{d(\tan x)}{dx}_{\left(\text{ at } x=\dfrac\pi4\right)}}{\dfrac{d(\sin x)}{dx}_{\left(\text{ at } x=\dfrac\pi4\right)}}$$
Following orion's way, I will set $\displaystyle\frac\pi4-x=2y$
$$\lim_{x\to\dfrac\pi4}\frac{1-\tan x}{1-\sqrt2\sin x}$$
$$=\lim_{y\to0}\frac{1-\tan\left(\dfrac\pi4-2y\right)}{1-\sqrt2\sin\left(\dfrac\pi4-2y\right)}$$
$$=\lim_{y\to0}\frac{1-\dfrac{1-\tan2y}{1+\tan2y}}{1-(\cos2y-\sin2y)}$$
Using double angle formula this becomes,
$$2\frac1{\lim_{y\to0}(1+\tan2y)\cos2y}\cdot 2\lim_{y\to0}\frac{\sin2y}{2y}\frac1{\lim_{y\to0}\dfrac{\sin y}y} \cdot\frac1{\lim_{y\to0}(\sin y+\cos y)}$$
Observe that all the limits reduce to $1$
First, convert it to limit around zero with: $$y=x-\frac{\pi}{4}$$ Expand all trigonometric functions with addition theorems. Things will become more obvious then. You can then use l'Hospital, further trigonometric manipulations to cancel things out, or just simple Taylor expansion.
EDIT:
$$\lim_{y\to 0}\frac{1-\tan(y+\pi/4)}{1-\sqrt{2}\sin(y+\pi/4)}$$ $$=\lim_{y\to 0}\frac{1-\frac{\tan y+\tan \pi/4}{1-\tan y \tan \pi/4}}{1-\sqrt{2}(\sin y\cos \frac{\pi}{4}+\cos y \sin{\frac{\pi}{4}})}$$ $$=\lim_{y\to 0}\frac{1-\frac{1+\tan y}{1-\tan y}}{1-(\sin y+\cos y)}$$ $$=\lim_{y\to 0}\frac{-2\tan y}{({1-\tan y})(1-(\sin y+\cos y))}$$ $$=\lim_{y\to 0}\frac{-2\tan y}{1-\sin y -\cos y-\tan y+(\sin^2 y/\cos y)+\sin y}$$ Multiply by cosine on both sides $$=\lim_{y\to 0}\frac{-2\sin y}{1+\cos y -2\cos^2 y-\sin y}$$ You can solve this with l'Hospital now: $$=\lim_{y\to 0}\frac{-2\cos y}{-\sin y -4\cos y\sin y-\cos y}=2$$
I oversimplified on purpose - you could do that a few lines earler. But it's nice to have it in this form in case you want to apply other methods (especially the Taylor approach).
