Delta Epsilon Proof of $\lim_{x\to\infty}\frac{\ln{x}^2}{x}=0$

Question

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.

Answer 1

... 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).

Answer 2

$\ln x<2\sqrt{x}$ so $$\left|\dfrac{\ln x^2}{x}\right|<\dfrac{4}{\sqrt{x}}<\dfrac{4}{\sqrt{M}}\leq\varepsilon$$ where $x>M$.