Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:$$\ln\left(\frac{4x}{2x}\right)=\ln(2)$$

share|cite|improve this question
This is unnecessary. What happens if you divide out a common factor of $x^2$ from top and bottom before taking the limit? – Semiclassical Jul 19 '14 at 0:48
The issue is rather the continuity of the natural log function at the limit of $\frac{2x^2+1}{x^2+1}$ as $x$ tends to $\infty$. – hardmath Jul 19 '14 at 0:49
I did not think to divide by a common factor. I see how that simplifies the problem leaving me with the same answer. – user2121620 Jul 19 '14 at 0:53
up vote 11 down vote accepted

For a continuous function $f$ on $a\in\overline{\Bbb R}$ we have $$\lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x))$$

share|cite|improve this answer
Only if $\lim \limits_{x\to a}(g(x))\in \text{dom}(f)$. – Git Gud Jul 19 '14 at 1:01
What's a proof for this? – YoTengoUnLCD Dec 20 '15 at 3:19

Yes, you can because $$\lim_{x\to\infty}f(g(x))=f(\lim_{x\to\infty}g(x))$$

But L'Hospital's is not necessary for this case as you can just factor out an $x^2$ from the numerator and denominator: $$\lim_{x\rightarrow \infty} \ln\left(\frac{2x^2+1}{x^2+1}\right)=\lim_{x\rightarrow \infty} \ln\left(\frac{2+\frac{1}{x^2}}{1+\frac{1}{x^2}}\right)$$

share|cite|improve this answer

Yes you can, because the logarithm function is continuous.

Remark: You really do not need L'Hospital's Rule to find the limit of $\frac{2x^2+1}{x^2+1}$. More concretely, divide top and bottom by $x^2$.

share|cite|improve this answer

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.

share|cite|improve this answer

In fact, we have the following proposition:

If $\lim_{x\to\infty}f(x)=a$ and $a>0$, then $\lim_{x\to\infty}\ln f(x)=\ln a$.

Now, $\lim_{x\to\infty}\frac{2x^2+1}{x^2+1}=\lim_{x\to\infty}\frac{4x}{2x}=2>0$. So


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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