TITLE: Am I allowed to apply L'Hospital's Rule inside of the natural logarithm function?
QUESTION [8 upvotes]: I have the following limit:
$$\lim_{x\rightarrow \infty} \ln\left(\frac{2x^2+1}{x^2+1}\right)$$
If I was finding the limit of only the terms inside the natural log function, I would have the indeterminate form: $$\frac{\infty}{\infty}$$
I want to know if I am allowed to apply L'Hospital's Rule to only the inside terms while ignoring the natural logarithm function, giving me the answer:$$\ln\left(\frac{4x}{2x}\right)=\ln(2)$$

REPLY [2 votes]: If you say that
$$
\lim_{x\rightarrow \infty} \ln\left(\frac{2x^2+1}{x^2+1}\right) = \ln\left( \lim_{x\to\infty}\frac{2x^2+1}{x^2+1} \right),
$$
that is not L'Hopital's rule.  That is continuity of the logarithm function.  You can apply L'Hopital's rule after that.  But L'Hopital's rule is overkill in this case, since the limit can easily be found be less powerful methods.