How do I solve $\lim\limits_{x\to\infty} \ \frac{\ln \ x}{x} = 0$? $$\lim\limits_{x\to\infty} \ \frac{\ln \ x}{x}  = 0$$
How can I know this without looking at the graph? What's the easiest way to equate that expression to 0?
I understand that an integer over x approaches 0 as x approaches 0, but since ln x is an increasing function as x approaches infinity, how do we know which effect outweighs the other?
 A: By L'Hospital rule
$$\lim_{x \rightarrow \infty}\frac{\ln x}{x}=\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{1}$$
$$\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{1}=\lim_{x \rightarrow \infty}\frac{1}{x}$$
$$\lim_{x \rightarrow \infty}\frac{1}{x}=0$$
A: Since $\ln t \le t$, for $t = \sqrt{x}$ we would have
$$\ln \sqrt{x}=\dfrac{1}{2}\ln x \leq \sqrt{x} \iff \ln x \leq 2\sqrt{x}$$
and, for $x>1$,
$$0\le\ln x\leq 2\sqrt{x}\iff 0\le\dfrac{\ln x}{x}\leq \dfrac{2}{\sqrt{x}}$$
With the squeeze theorem we have $\dfrac{\ln x}{x} \underset{x\to +\infty}{\longrightarrow}0$
A: $$\lim_{x\to \infty} {\ln x \over x} =\lim_{x\to \infty} \ln x^{1/x} = \ln\left( \lim_{x\to \infty}  x^{1/x}\right) = \ln 1 = 0$$
Limit used : How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$?
A: With $x=e^t$ we have to investigate $\lim_{t\to \infty} \frac{t}{e^t} $. For $t>0$ we have
$0<\frac{t}{e^t}=\frac{t}{1+t+\frac{t^2}{2}+...} \le \frac{t}{\frac {t^2}{2}}=\frac{2}{t}$.
Your turn !
A: A high school proof, using only basic properties of the integral:
For $t\ge 1$, we have  $\sqrt t\le t$, so $\dfrac1t\le\dfrac 1{\sqrt t}$ and hence, if $x\ge 1$,
$$ \ln x=\int_1^x\frac{\mathrm d\mkern1mut}{t}\le\int_1^x\frac{\mathrm d\mkern1mut}{\sqrt t}=2(\sqrt x-1)<2\sqrt x$$
from which we deduce at once
$$0\le\frac{\ln x}{x}<\frac{2\sqrt x}{x},\quad\text{which tends to }0.$$
