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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.

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up vote 2 down vote accepted

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 $\lim_{y\to 0}{y\over \log(1+y)}=\lim_{y\to 0}{1\over{\log(1+y)\over y}}=1$

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Yes. I tried this way, too. but what we have to do next? – KiberPrestupnik Oct 18 '13 at 9:17
    
Understood. Thanks a lot for your help! – KiberPrestupnik Oct 18 '13 at 9:18
1  
you are welcome – La Belle Noiseuse Oct 18 '13 at 9:22

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