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

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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 at 0:48
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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 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 at 0:53

5 Answers 5

up vote 9 down vote accepted

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

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Only if $\lim \limits_{x\to a}(g(x))\in \text{dom}(f)$. –  Git Gud Jul 19 at 1:01

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: $$\lim_{x\rightarrow \infty} \ln\left(\frac{2x^2+1}{x^2+1}\right)=\lim_{x\rightarrow \infty} \ln\left(\frac{2+\frac{1}{x^2}}{1+\frac{1}{x^2}}\right)$$

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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 $x^2$.

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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

$$\lim_{x\to\infty}\ln\left(\frac{2x^2+1}{x^2+1}\right)=\ln\lim_{x\to\infty}\left(\frac{2x^2+1}{x^2+1}\right)=\ln2.$$

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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.

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