0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.