Why is $\lim_{x \to \infty} \frac{\ln x}{x^2} = 0$?

Question

The question is pretty straightforward, I just can't seem to figure it out. Shouldn't $\ln(\infty) =\infty$, and the function would be $\frac{\infty}{\infty}$? Is it because of that rule that says that an exponential function grows faster than the others?

I'm very confused and I'm sorry if I don't make any sense and for the lack of styling. By the way, I have to solve without using l'Hopital's rule.

Answer 1

Note that $\ln(x) \leq x$ for all $x>1$, and certainly, $$\lim_{x\rightarrow \infty} \frac{1}{x} = 0.$$ You also need to state that the log is positive for $x$ sufficiently large to conclude that the limit is zero.

Answer 2

It is in the $\frac{\infty}{\infty}$ indeterminate form, but unlike exponentials, logarithms grow much more slowly than power functions. If you look at the graph of a logarithm, it almost looks like it has a horizontal asmyptote, but it doesn't. It continues to grow to $\infty$, but at an exceedingly slow rate.

If you prefer to think in exponentials, you could make a variable substitution. Let $u = \ln x$, or in other words, $x = e^u$. As $x \to \infty$, we also have $u \to \infty$, so

$$\lim_{x \to \infty} \frac{\ln x}{x^2} = \lim_{u \to \infty} \frac{u}{(e^u)^2} = \lim_{u \to \infty} \frac{u}{e^{2u}}.$$ As you seem to know, exponentials grow much faster than power functions, so this limit will be $0$.

Answer 3

$$ \lim_{x \to \infty} \frac{\ln x}{x^2} = 0. $$ We must make a first substitution as $x=\frac{1}{y}$, therefore, the next expression will appear: $$ \lim_{y\to{0}}{(-y^{2}\ln(y))}. $$ Now we should make the second substitution as $y=(w+1)$, and, therefore, the next expression will be received: $$ \lim_{w\to{-1}}{(-(w+1)^{2}\ln(w+1))}. $$ After the last transformation I strongly recommend to use the Taylor series relatively $\ln(w+1)$: $$ \ln(w+1)=w-\frac{w^{2}}{2}+\frac{w^{3}}{3}-....\tag{1} $$ Finally we will obtain, that: $$ \lim_{w\to{-1}}{(-(w+1)^{2}\ln(w+1))}=\lim_{w\to{-1}}{\left(-(w+1)^{2}\left(w-\frac{w^{2}}{2}+\frac{w^{3}}{3}-....\right)\right)}= \\ \lim_{w\to{-1}}{\left({0}\cdot\left(w-\frac{w^{2}}{2}+\frac{w^{3}}{3}-....\right)\right)}=\lim_{w\to{-1}}{0}=0. $$ Therefore, $$ \lim_{x \to \infty} \frac{\ln x}{x^2} = 0. $$