Yahir Tucker

Answered

2022-06-26

How to prove that $\mathrm{ln}(x)<x$ for $x\to \mathrm{\infty}$

During my calculus homework I need to prove some limits without using L'Hôpital's rule. I have difficulties to show rigorously that $\mathrm{ln}(x)<x$ for big enough x.

For example, I need to find the image of the continues function $f:R\to R$ such that for every $x\in R$: $|f(x)-x{e}^{\sqrt{\left|x\right|}}|<{x}^{4}$

I've tried to prove that $x{e}^{\sqrt{\left|x\right|}}-{x}^{4}\to \mathrm{\infty}$ if $x\to \mathrm{\infty}$, and then the image of $f$ will be all the reals. Unfortunately, I don't find the way to make it formal enough.

During my calculus homework I need to prove some limits without using L'Hôpital's rule. I have difficulties to show rigorously that $\mathrm{ln}(x)<x$ for big enough x.

For example, I need to find the image of the continues function $f:R\to R$ such that for every $x\in R$: $|f(x)-x{e}^{\sqrt{\left|x\right|}}|<{x}^{4}$

I've tried to prove that $x{e}^{\sqrt{\left|x\right|}}-{x}^{4}\to \mathrm{\infty}$ if $x\to \mathrm{\infty}$, and then the image of $f$ will be all the reals. Unfortunately, I don't find the way to make it formal enough.

Answer & Explanation

popman14ee

Expert

2022-06-27Added 19 answers

For the first part:

Note the following:

1.$\mathrm{log}(1)=0<1$

2.${\mathrm{log}}^{\prime}(x)=\frac{1}{x}\le 1=\frac{d}{dx}x$ for $x\ge 1$

3.${\mathrm{log}}^{\prime}(x)=\frac{1}{x}\ge 1=\frac{d}{dx}x$ for $x\le 1$

So you can integrate over an appropriate interval to get $\mathrm{log}(x)<x$ for all $x$.

For the second part:

Try to use

$\underset{x\to \mathrm{\infty}}{lim}\frac{{e}^{x}}{{x}^{n}}=\mathrm{\infty}$

for any natural number $n$.

Note the following:

1.$\mathrm{log}(1)=0<1$

2.${\mathrm{log}}^{\prime}(x)=\frac{1}{x}\le 1=\frac{d}{dx}x$ for $x\ge 1$

3.${\mathrm{log}}^{\prime}(x)=\frac{1}{x}\ge 1=\frac{d}{dx}x$ for $x\le 1$

So you can integrate over an appropriate interval to get $\mathrm{log}(x)<x$ for all $x$.

For the second part:

Try to use

$\underset{x\to \mathrm{\infty}}{lim}\frac{{e}^{x}}{{x}^{n}}=\mathrm{\infty}$

for any natural number $n$.

preityk7t

Expert

2022-06-28Added 6 answers

Hint: Consider the function $f(x)=\mathrm{ln}(x)-x$. Note that $f(1)<0$ and show this function is monotonically decreasing (from $x=1$) by taking the derivative

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