sedeln5w
2022-06-11
Answered

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

Answered 2022-06-12
Author has **12** answers

$\mathrm{ln}x=2\mathrm{ln}{x}^{\frac{1}{2}}\le 2\sqrt{x}$ (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.

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.

Hailie Blevins

Answered 2022-06-13
Author has **8** answers

Extending the approximation

${e}^{x}\ge 1+x$

into

${e}^{x}\ge 1+x+\frac{{x}^{2}}{2}>\frac{(x+1{)}^{2}}{2}$

and

${e}^{x}\ge 1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{6}>\frac{(x+1{)}^{3}}{6}$

allows

${e}^{x}>\frac{(x+1{)}^{n}}{n!}$

for all $n\in \mathbb{Z}$

So,

$x\ge n\mathrm{ln}(1+x)-\mathrm{ln}n!\ge n\mathrm{ln}x-\mathrm{ln}n!$

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.

${e}^{x}\ge 1+x$

into

${e}^{x}\ge 1+x+\frac{{x}^{2}}{2}>\frac{(x+1{)}^{2}}{2}$

and

${e}^{x}\ge 1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{6}>\frac{(x+1{)}^{3}}{6}$

allows

${e}^{x}>\frac{(x+1{)}^{n}}{n!}$

for all $n\in \mathbb{Z}$

So,

$x\ge n\mathrm{ln}(1+x)-\mathrm{ln}n!\ge n\mathrm{ln}x-\mathrm{ln}n!$

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.

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