Is this way of finding $\lim\limits_{x\to +\infty}(x-\ln(x^2+1))$ valid? I needed to find: $$\lim\limits_{x\to +\infty}(x-\ln(x^2+1))$$
So here are the steps I took:
Step 1: Replace $x$ with $\ln(e^x)$: 
$$\lim\limits_{x\to +\infty}\left(\ln(e^x)-\ln(x^2+1)\right)$$
$$\lim\limits_{x\to +\infty}\ln\left(\frac{e^x}{x^2+1}\right)$$
Step 2: Bring the limit inside of the natural log function since it is continuous on the required interval.
$$\ln\left(\lim_{x\to +\infty}\frac{e^x}{x^2+1}\right)$$
Step 3: Apply L'Hospital's rule twice and evaluate:
$$\ln\left(\lim_{x\to +\infty}e^x\right)$$
$$\ln(+\infty) = +\infty$$
My question is whether step 2 is valid here because $\lim\limits_{x\to \infty}\frac{e^x}{x^2 + 1}$ doesn't exist (its $+\infty$), and in order to move the limit operator inside the function the limit $\lim\limits_{x\to \infty}\frac{e^x}{x^2 + 1}$ must exist according to this theorem in a book about Calculus (ISBN 978-0-470-64769-1):
If it's not valid, what would be a valid way to find the limit? 
 A: Let $M>0$ be given. Since $\ln(u)\to \infty$ as $u\to\infty$, there exists $K$ such that 
for all $x$ with $\frac{e^x}{x^2+1}>K$ it follows that $\ln\left(\frac{e^x}{x^2+1}\right)>M$. Since $\frac{e^x}{x^2+1}\to \infty$ as $x\to \infty$, there exists $N$ such that 
$$
x>N\implies \frac{e^x}{x^2+1}>K\implies \ln\left(\frac{e^x}{x^2+1}\right)>M.
$$
By definition of a limit it follows that 
$$
\lim_{x\to\infty}\frac{e^x}{x^2+1}=\infty.
$$
Note we can mimic  the same argument to conclude that if $f(x)\to \infty$ as $x\to \infty$ and $g(x)\to \infty$ as $x\to \infty$, then $g(f(x))\to \infty$ as $x\to \infty$.
A: In step 2 all you need is the fact that $\ln (y) \to \infty$ as $y\to \infty$. Since $\frac {e^{x}} {x^{2}+1} \to \infty$ it follows that $\ln (\frac {e^{x}} {x^{2}+1}) \to \infty$. To prove that $\ln (y) \to \infty$ as $y\to \infty$ assume that this is false. Since $\ln \, x$is an increasing function, if it doesn't not tend to infinity, it would be bounded for $x>1$, say $\ln\, x <C$ for all $x>1$. You get a contradiction from this if you take $x=e^{C}$.
