# I wanted to prove <munder> <mo form="prefix">lim <mrow class="MJX-TeXAtom-ORD"> x

I wanted to prove $\underset{x\to \mathrm{\infty }}{lim}\frac{\mathrm{ln}x}{x}=0$ by the squeeze theorem.
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hofyonlines5
$\mathrm{ln}x=2\mathrm{ln}{x}^{\frac{1}{2}}\le 2\sqrt{x}$ (using the inequality $y\ge \mathrm{ln}y$ for $y>0$ that you obtained.)
So it follows that: for all $x>0$
$|\frac{\mathrm{ln}x}{x}|\le 2\frac{\sqrt{x}}{x}$
It follows by squeeze theorem that $\underset{x\to \mathrm{\infty }}{lim}|\frac{\mathrm{ln}x}{x}|=0.$ Now note that $-|\frac{\mathrm{ln}x}{x}|\le \frac{\mathrm{ln}x}{x}\le |\frac{\mathrm{ln}x}{x}|$ and hence the result follows by squeeze theorem.
###### Not exactly what you’re looking for?
Hailie Blevins
Extending the approximation
${e}^{x}\ge 1+x$
into
${e}^{x}\ge 1+x+\frac{{x}^{2}}{2}>\frac{\left(x+1{\right)}^{2}}{2}$
and
${e}^{x}\ge 1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{6}>\frac{\left(x+1{\right)}^{3}}{6}$
allows
${e}^{x}>\frac{\left(x+1{\right)}^{n}}{n!}$
for all $n\in \mathbb{Z}$
So,
$x\ge n\mathrm{ln}\left(1+x\right)-\mathrm{ln}n!\ge n\mathrm{ln}x-\mathrm{ln}n!$
So
$\frac{x}{n}+\frac{\mathrm{ln}n!}{n}\ge \mathrm{ln}x$
$\frac{1}{n}+\frac{\mathrm{ln}n!}{nx}\ge \frac{\mathrm{ln}x}{x}$
As we can make both terms on the LHS arbitrarily small, we are done.