Is this way of finding $\lim\limits_{x\to +\infty}(x-\ln(x^2+1))$ valid?

Question

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?

Answer 1

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 $x\to \infty$.

Answer 2

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 $x=e^{C}$.