How to prove that $\lim_{x \to 0^{+}} \frac{e^{-1/x}}{x} = 0$

Question

Today someone asked me how to calculate $\lim_{x \to 0^{+}} \frac{e^{-1/x}}{x}$. At first sight that limit is $0$, because the exponential decreases faster than the lineal term in the denominator. However, I didn't know how to prove it formally.

I thought of expressing the exponential with its series expansion, but the person who asked me this didn't know about series expansions yet. Niether does he know about L'Hopital.

So, is there a "simple" way to prove that $\lim_{x \to 0^{+}} \frac{e^{-1/x}}{x}=0$?

Answer 1

You can use that, by continuity

$$\lim_{x\to 0^+}e^{-\frac1x}\cdot\frac{1}{x} = \exp\left(\lim_{x\to 0^+}\ln\left(e^{-\frac1x}\cdot\frac{1}{x}\right)\right)$$

The limit inside is (correcting as per the comments)

$$\lim_{x\to 0^+}\ln\left(e^{-\frac1x}\right)+\ln\left(\frac1x\right)=\lim_{x\to 0^+}-\left(\frac1x + \ln(x)\right)$$

so we need only show the expression in parenthesis goes to $+\infty$. Write the expression as $$\frac{1+x\ln(x)}{x}=\frac{1+\ln(x^x)}{x}$$

and as $x \to 0^{+}$, $x^x$ approaches $1$, so the numerator of the expression approaches $1$. It follows that as $x \to 0^+$, the expression approaches $+\infty$ as desired.

Answer 2

Here's a proof that relies on the fundamental theorem of calculus and the fact that $\frac{d}{dx}e^x=e^x$, together with the fact that $f(x)=e^x$ is positive and increasing:

First set $y=\frac{1}{x}$, so that the limit becomes $$ \lim_{y\to\infty}ye^{-y} $$

Next, if $y\geq 0$ then $$ e^y-1=\int_0^ye^t\;dt\geq \int_0^y\;dt=y $$ since $e^t\geq e^0=1$ for all $t\geq 0$. Thus $e^t\geq 1+t$, so integrating again shows that $$ e^y-1=\int_0^ye^t\;dt\geq \int_0^y1+t\;dt=y+\frac{y^2}{2} $$ so $e^y\geq 1+y+\frac{y^2}{2}$ for all $y\geq 0$.

Therefore $$ 0\leq ye^{-y}\leq \frac{y}{1+y+\frac{y^2}{2}}$$ for all $y\geq 0$, so we can conclude that $\lim_{y\to\infty}ye^{-y}=0$ by the squeeze theorem.

Answer 3

This is equivalent to showing that for $A> 1$ $$ \frac{y}{A^y} \rightarrow 0 $$ as $y \rightarrow \infty$.

You can see that pretty clearly if you pick an $n$ such that $A^n > 2$ and let $y$ grow through multiples of $n$.

Answer 4

I'll use a constant $c>1$ instead of $e$. Since the expression is nonnegative to the right-hand side of zero, we need to show that for all $\epsilon>0$, there exists $\delta>0$ such that for all $0<x<\delta$, $$ \frac{c^{-1/x}}{x}<\epsilon. $$ Moving some terms around, the above holds if and only if $$ \left(\epsilon x\right)^{x}>1/c. $$ Note that as $x\searrow 0$, $\left(\epsilon x\right)^{x}\rightarrow1$. Therefore, by continuity, we can find a $\delta$ as prescribed above, since $1/c<1$.

Answer 5

Putting $t = 1/x$ we see that $t \to \infty$ as $x \to 0^{+}$. Also the function is transformed into $t/e^{t}$. Next we put $e^{t} = y$ so that $y \to \infty$ as $t \to \infty$. Thus the function is transformed into $(\log y)/y$. Since $y \to \infty$ we can assume $y > 1$ so that $\sqrt{y} > 1$. We have the inequality $$\log x \leq x - 1$$ for all $x \geq 1$. Replacing $x$ by $\sqrt{y}$ we get $$0 \leq \log\sqrt{y} \leq \sqrt{y} - 1 < \sqrt{y}$$ or $$0 \leq \log y < 2\sqrt{y}$$ or $$0 \leq \frac{\log y}{y} < \frac{2}{\sqrt{y}}$$ for $y > 1$. Applying squeeze theorem when $y \to \infty$ we get $$\lim_{y \to \infty}\frac{\log y}{y} = 0$$ and therefore the desired limit is $0$.

Answer 6

Suppose all we know about $e$ is that it's bigger than $1$. To begin, note that, for $x\gt0$, we have

$$0\lt{e^{-1/x}\over x}={1/x\over e^{1/x}}\lt{\lceil1/x\rceil\over e^{\lfloor1/x\rfloor}}\le e{\lceil1/x\rceil\over e^{\lceil1/x\rceil}}$$

This allows us to switch from the limit of the continuous variable $x$ going to $0$ to the limit of a discrete variable, $n=\lceil1/x\rceil$, going to $\infty$. In other words, we need only show that

$$\lim_{n\to\infty}a_n=0\quad\text{where}\quad a_n={n\over e^n}$$

It's easy to see that

$${a_{n+1}\over a_n}={n+1\over ne}\lt1\quad\text{if}\quad {1\over e-1}\lt n$$

and thus the infinite sequence is eventually decreasing. The completeness property of the real numbers tells us that a decreasing sequence that's bounded below (in this case by $0$) has a limit. So we now know that

$$\lim_{x\to0^+}{e^{-1/x}\over x}=\lim_{n\to\infty}{n\over e^n}=L$$

for some real number $L$. We need to show that $L=0$. There are various ways to do this. One I like is to use the fact that any subsequence of a convergent sequence converges to the same limit, followed by the fact that the limit of a product is the product of the limits (when all the limits exist). Thus

$$L=\lim_{n\to\infty}{2n\over e^{2n}}=\lim_{n\to\infty}\left({2\over n}\cdot{n\over e^n}\cdot{n\over e^n}\right)=\lim_{n\to\infty}\left({2\over n}\right)\lim_{n\to\infty}\left({n\over e^n}\right)\lim_{n\to\infty}\left({n\over e^n}\right)=0\cdot L\cdot L=0$$