Evaluate $\lim\limits_{w\rightarrow\infty}\frac{\ln{w^2}}{\ln{w^3}+1}$
This particular limit is giving me trouble. Am I correct that $\lim\limits_{x\rightarrow a}(\ln{f(x)})=\ln{\lim\limits_{x\rightarrow a}}f(x)$? That is the approach I took but it didn't seem to help. I must have a hole in my understanding here because I can't figure out how to proceed. Any hints or advice would be greatly appreciated
Unfortunately not if $\lim_{x\to a} f(x) = \infty$ which is the case here.
$\ln w^k = k\ln w$ so replace $\ln w$ with $v$ and you have
$\lim_{w\to \infty}\frac {\ln w^2}{\ln w^3 + 1} =\lim_{w\to \infty} \frac {2\ln w}{3\ln w^3 + 1}=\lim_{e^v\to \infty}\frac {2v}{3v + 1}$
Can you do that?