Prove that $\frac{x}{e^x}$ tends to zero as $x \to \infty $

Question

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!

Answer 1

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 $$0<\frac{x}{e^{x^2}}<\frac1x.$$

Answer 2

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 $$\frac{x}{e^{x^2}}\to 0$$

Answer 3

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 $x\in (n,n+1)$

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

Answer 4

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 $$ \frac{x}{e^x}\le2e^{-x/2}\tag3 $$

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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, $$ \begin{align} \frac{x}{e^x} &\le\frac{x}{\left(1+\frac{x}2\right)^2}\\[3pt] &\le\frac4x\tag5 \end{align} $$

Answer 5

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)}} $.