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:$$\ln\left(\frac{4x}{2x}\right)=\ln(2)$$
On
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 $x^2$.
On
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: $$\lim_{x\rightarrow \infty} \ln\left(\frac{2x^2+1}{x^2+1}\right)=\lim_{x\rightarrow \infty} \ln\left(\frac{2+\frac{1}{x^2}}{1+\frac{1}{x^2}}\right)$$
On
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
$$\lim_{x\to\infty}\ln\left(\frac{2x^2+1}{x^2+1}\right)=\ln\lim_{x\to\infty}\left(\frac{2x^2+1}{x^2+1}\right)=\ln2.$$
On
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.
For a continuous function $f$ on $a\in\overline{\Bbb R}$ we have $$\lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x))$$