I'm supposed to evaluate this limit using L'Hopital's rule.
$$ \lim_{x \to 0} \frac {\ln(\tan 2x )}{\ln(\tan 3x )} $$
I find the indeterminate form of $\frac{0}{0}$. The latter tells me that L'Hopital's is an option, after applying the rule once I end up with:
$$ \lim_{x \to 0} \frac {\frac {1}{\tan 2x}*\sec^2(2x)*2}{\frac {1}{\tan 3x}*\sec^2(3x)*3} $$
After this step however I seem to get lost in my own translation time and time again. Can someone point me in the right direction?
Best Answer
HINT:
Note that
$$\frac{\tan(3x)}{\tan(2x)}=\left(\frac{\sin(3x)}{3x}\right)\left(\frac{3x}{2x}\right)\left(\frac{2x}{\sin(2x)}\right)\left(\frac{\sec(3x)}{\sec(2x)}\right)$$