calculus - How do I solve this limit? $lim_{x to frac{pi}{4}}frac{sin x-cos x}{ln(tan x)}$

I can't really see the right way to solve this limit. My attempt is:

$$\lim_{x \to \frac{\pi}{4}}\frac{\sin x-\cos x}{\ln(\tan x)}=\left(\lim_{x \to \frac{\pi}{4}}\frac{\sin x-\cos x}{\ln(\tan x)}\right):\cos x = \lim_{x \to \frac{\pi}{4}}\frac{\tan x-1}{\frac{\ln(\tan x)}{\cos x}} = \frac{0}{\frac{0\cdot{2}}{0\cdot\sqrt{2}}}$$ So this answer is wrong.

But I would like to understand how to solve this problem without using derivation or L'hôpital's rule.

Answer

Hint: your expression can be transformed as $\tan x-1\over \log (1+\tan x -1)$

Now take $y=\tan x -1$ so $y\to 0$ as $x\to{\pi\over 4}$

so now in $y$ variable $\lim_{y\to 0}{y\over \log(1+y)}=\lim_{y\to 0}{1\over{\log(1+y)\over y}}=1$