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As the title states, I want to prove $$\lim_{x \to \infty} \frac{x}{e^x} =0$$

Clearly, L'Hopital's rule easily solves this. However, I'm curious to see if there's another way to prove it, without involving some differential or integral calculus (that is, by algebraic means). What I'm really interested about, is to prove that $$\lim_{x \to \infty} \frac{x}{e^{x^2}}=0 $$ I assume that proving the first limit will provide a way to prove the second one, using the squeeze method. If you know a direct way to prove the second limit, it will be more than perfect.

Thanks in advance!

Abraham Zhang
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NotAMathematician
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  • An explicit way to prove it would be via the $\epsilon-\delta$ definition of limit. Given $\epsilon\gt 0$, can you find a $\delta\gt 0$ such that for all $x\gt\delta$, we have $|x/e^x|\lt\epsilon$ ? – Prasun Biswas Mar 22 '18 at 01:37
  • Another one: https://math.stackexchange.com/q/873373 – Martin R Mar 22 '18 at 08:10
  • @NotAMathematician Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 24 '18 at 21:19

5 Answers5

4

For $x>0$, $e^x=1+x+x^2/2+\cdots>x^2/2$ so $$0<\frac{x}{e^x}<\frac2x.$$ Likewise $e^{x^2}=1+x^2+\cdots>x^2$ so $$0<\frac{x}{e^{x^2}}<\frac1x.$$

Angina Seng
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1

Mean Value Theorem

By the Mean Value Theorem, there is a $\xi\in(x/2,x)$ so that $$ \begin{align} \frac{e^x-e^{x/2}}{x-x/2} &=e^{\xi}\\ &\ge e^{x/2}\tag1 \end{align} $$ However, we also have $$ \begin{align} \frac{e^x-e^{x/2}}{x-x/2}\le2\,\frac{e^x}{x} \end{align}\tag2 $$ Putting $(1)$ and $(2)$ together, we get $$ \frac{x}{e^x}\le2e^{-x/2}\tag3 $$

Bernoulli's Inequality

Bernoulli's Inequality implies that $$ \begin{align} \left(1+\frac{x}n\right)^n &\le\lim_{n\to\infty}\left(1+\frac{x}n\right)^n\\[6pt] &=e^x\tag4 \end{align} $$ Therefore, $$ \begin{align} \frac{x}{e^x} &\le\frac{x}{\left(1+\frac{x}2\right)^2}\\[3pt] &\le\frac4x\tag5 \end{align} $$

robjohn
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For the sequence by ratio test

$$a_n=\frac{n+1}{e^n}\implies \frac{a_{n+1}}{a_n}=\frac{n+2}{e^{n+1}}\frac{e^n}{n+1}=\frac{n+2}{n+1}\frac1e\to \frac1e<1\implies \frac{n+1}{e^n}\to 0$$

then observe that for $x\in (n,n+1)$

$$0<\frac{x}{e^x}\le\frac{n+1}{e^n} \to0$$

user
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This is equivalent to $\dfrac{\ln(x)}{x} \to 0$ and this has been proved many times here, often by me.

One easy proof:

$\begin{array}\\ \ln(x) &=\int_1^x \dfrac{dt}{t}\\ &<\int_1^x \dfrac{dt}{t^c} \qquad\text{where } 0 < c < 1\\ &=\int_1^x t^{-c}dt\\ &=\dfrac{t^{1-c}}{1-c}|_1^x\\ &=\dfrac{x^{1-c}-1}{1-c}\\ &<\dfrac{x^{1-c}}{1-c}\\ \end{array} $

Now choose $c$ appropriately. For example, $c=\frac12$ gives $\ln(x) \lt 2x^{1/2}$ so $\dfrac{\ln(x)}{x} \lt \dfrac{2}{x^{1/2}} $.

Setting $x = e^y$ gives $\dfrac{y}{e^y} \lt \dfrac{2}{e^{y/2}} $.

If $c = 1-d$ where $d$ is small, this is $\ln(x) \lt\dfrac{x^{d}}{d} $ or $\dfrac{\ln(x)}{x} \lt \dfrac1{dx^{1-d}} $.

In particular, if $d = 1/n$, this is $\dfrac{\ln(x)}{x} \lt \dfrac{n}{x^{1-1/n}} $ or $\dfrac{y}{e^y} \lt \dfrac{n}{e^{y(1-1/n)}} $.

marty cohen
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First show $$\frac{x}{a^x}\to 0$$ for any $a>1$. This means $$\frac{x}{e^{2x}}\to 0$$ and write this as $$\frac{x^2}{e^{2x^2}}\to 0$$ now take the square root to get $$\frac{x}{e^{x^2}}\to 0$$

Rene Schipperus
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