TITLE: Delta Epsilon Proof of $\lim_{x\to\infty}\frac{\ln{x}^2}{x}=0$
QUESTION [6 upvotes]: I am trying to prove that $\lim_{x\to\infty}\frac{\ln{x}^2}{x}=0$ via the delta epsilon definition of a limit but I keep getting hung up.
I have started the proof like this:
$\forall \epsilon>0, \exists \delta$ so if $\delta>0$, $|\frac{\ln{x}^2}{x}-0|<\epsilon$. 
If we suppose that $|\frac{\ln{x}^2}{x}-0|<\epsilon$ and solve for x we get...
When I try solving the inequality for $x$, it keeps cancelling.
What I have tried:
$2x^{-1}\ln{x}<\ln{\epsilon}$
Which becomes:
$2x^{-1}(x^1)<e^{\ln{\epsilon}}$
But then, $x^{-1}$ and $x^1$ multiply to $x^0$, which is $1$ and $x$ is removed from the problem. 
Any pointers as how to proceed? I know that the limit is $0$, I just can't seem to get past solving for $x$.
Thank you.

REPLY [2 votes]: ... I have started the proof like this:
$\forall \epsilon>0, \exists \delta$ so if $\delta>0$, $|\frac{\ln{x}^2}{x}-0|<\epsilon$.

The $\delta$ in your attempt has nothing to do with $\epsilon$.
By definition, given $\epsilon>0$, you need to show that there exists $M>0$ such that
$$
x>M \quad \textbf{implies }\quad |\frac{2\ln x}{x}|<\epsilon\tag{1}
$$
Now for $x>1$,
$$
\ln x=\int_1^x\frac{1}{t}\ dt\leq \int_1^x\frac{1}{\sqrt{t}}\ dt=2\sqrt{t}\big|_1^x
< 2\sqrt{x},
$$
which implies that
$$
\frac{2\ln x}{x}<\frac{4\sqrt{x}}{x}=\frac{4}{\sqrt{x}},\quad x>1.
$$
So you want
$
\frac{4}{\sqrt{x}}<\epsilon,
$
which is true if
$$
x>(4/\epsilon)^2.
$$
Now, setting
$M=(4/\epsilon)^2$, you have the implication (1).