Functions and limits?

If $2f(\sin x) + \sqrt{2}f(\cos x) = \tan x$ then evaluate :
$\lim_{x\rightarrow 1} \sqrt{(1 - x)}f(x)$
Rewrite the condition as: $2f(a)+\sqrt2f(b)=\frac ab$ and $2f(b)+\sqrt2f(a)=\frac ba$ for all $a^2+b^2=1$. Using a 2x2 matrix to solve, we get $f(a)=\frac ab-\frac{\sqrt2}2\frac ba$ and $f(b)=\frac ba-\frac{\sqrt2}2\frac ab$, so $f(\sin x)=\tan x-\frac{\sqrt2}2\cot x$ and $f(\cos x)=\cot x-\frac{\sqrt2}2\tan x$. Rewriting another way, $f(x)=\frac x{\sqrt{1-x^2}}-\frac{\sqrt2}2\frac{\sqrt{1-x^2}}x$
So $\lim_{x\to 1}\sqrt{1-x}f(x)=\lim_{x\to1}\frac x{\sqrt{1+x}}-\frac{\sqrt2}2\frac{\sqrt{1+x}(1-x)}x=\frac{\sqrt2}2$