Trigonometric limit $\lim_{x\to\pi/4}\frac{1-\tan x}{1-\sqrt{2}\sin x}$ The limit is
$$\lim_{x\to\pi/4}\frac{1-\tan x}{1-\sqrt{2}\sin x}$$ I was able to solve it using L'hopital and the answer that I got was $2$.

Can you please confirm if the answer is right and suggest some other way to evaluate the limit without using L'hopital?
Multiply both deminator and numerator to the $\left( 1+\sqrt { 2 } \sin { x }  \right) $ and $\left( 1+\tan { x }  \right)  $ respectively
\begin{align*}
\lim_{x\to\frac{\pi}{4}}{\frac{1-\tan x}{1-\sqrt{2}\sin x}}={}&amp;\lim_{x\to\frac{\pi}{4}}{\frac{\left(1-\tan^2\!x\right)\left(1+\sqrt2\sin x\right)}{\left(1-2\sin^2\!x\right)\left(1+\tan x\right)}}={} \\
{}={}&amp;\lim_{x\to\frac{\pi}{4}}{\frac{\frac{\cos(2x)}{\cos^2\!x}\left(1+\sqrt2\sin x\right)}{\cos(2x)\left(1+\tan x\right)}}={} \\
{}={}&amp;\lim_{x\to\frac{\pi}{4}}\frac{\left(1+\sqrt2\sin x\right)}{\cos^2\!x\left(1+\tan x\right)}=2.
\end{align*}
Let $f(x) = 1-\tan x, g(x) = 1-\sqrt 2\sin x.$ The expression equals

$$\frac{f(x) - f(\pi/4)}{g(x) - g(\pi/4)} = \frac{(f(x) - f(\pi/4))/(x-\pi/4)}{g(x) - g(\pi/4)/(x-\pi/4)} .$$

By definition of the derivative, this $\to f'(\pi/4)/g'(\pi/4)$ as $x\to \pi/4.$ This quotient of derivatives is easy to compute.