<s>[INST] 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?
 [/INST] 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.
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