# Proof of limit of exponential function

I was doing a proof that for $$a\in(0,1)$$ $$\lim_{x\to\infty}a^x=0$$ Proof For arbitrary $$\epsilon>0$$ we wish to find $$x_0\in \mathbb{R}$$ such that $$\forall x\in \mathbb{R}$$ if $$x>x_0$$ then $$|a^x|<\epsilon$$. By some analysis we set $$x_0= \frac{\ln{\epsilon}}{\ln{a}}$$. Now, if we take $$x>x_0$$ we have \begin{align*} x>\frac{\ln{\epsilon}}{\ln{a}}\\ x\ln{a}<\ln{\epsilon}\\ \ln{a^x}<\ln{\epsilon}\\ a^x=|a^x|<\epsilon \end{align*} Here i saw that the fact that $$a\in (0,1)$$ helps to convert the $$<$$ to $$>$$.

Now, obviously for $$a>1$$ this shouldn't hold, so in this case i wanted to show that $$\lim_{x\to\infty}a^x\neq0$$ that is there exists $$\epsilon >0$$ such that for any $$x_0\in \mathbb{R}$$ there is $$x\in \mathbb{R}$$ we have both $$x>x_0$$ and $$a^x\geq \epsilon$$. Now, i have no idea what should i pick, it seems reasonable to me that $$\epsilon$$ should somehow depend on $$x_0$$ but I can't do that according to the order of the quantifiers. So, maybe take $$\epsilon = 1/2$$. Now consider arbitrary $$x_0$$, we wish to find some $$x$$ such that if $$x>x_0$$ we get that $$a^x\geq \epsilon$$. We know that $$a^x$$ is always positive. Take $$x=x_0+1$$ to obtain $$a^x=a^{x_0+1}=a^{x_0}+a\geq 0+a>1>1/2>\epsilon$$ I am wondering, if this is the correct approach and possible if there can be something more elegant done? Thanks

• I think your proof is perfect :) – Mike Earnest Sep 28 '18 at 19:24

## 4 Answers

Your approach is very nice! Here is a different argument to show the same result.

Let $$x_0 = \sup\{a\in\mathbb R\;|\;\exists x\in\mathbb N\quad a^x\leq\frac12\}$$Where $$\sup$$ is the supremum. It's obvious that $$x_0\geq\frac12$$. Our goal should be to show that $$x_0\geq1$$ (in fact, $$x_0=1$$, but we don't need to prove this for this problem).

We prove this by proof of contradiction. Suppose $$x_0<1$$. We then know that $$x_0<\sqrt{x_0}<1$$, so pick $$k$$ such that $$x_0. Hence, $$k^2.

By the property of $$\sup$$, we can find a $$l$$ such that $$k and $$l\in\{a\in\mathbb R\;|\;\exists x\in\mathbb N\quad a^x\leq\frac12\}$$. So, $$\exists n\in\mathbb N$$ such that $$l^n<\frac12$$. But, $$k^{2n}So, $$k\in\{a\in\mathbb R\;|\;\exists x\in\mathbb N\quad a^x\leq\frac12\}$$, a contradiction of the supremum property.

Hence, $$x_0\geq1$$. And since $$x^n$$ is a strictly increasing function on non-negative numbers, we've shown that $$\forall x\in[0,1],\;\exists n\in\mathbb N\quad x^n\leq\frac12$$This is equivalent to the theorem we seek to show, since $$\lim_{n\to\infty}\bigg(\frac12\bigg)^n=0$$

By the Binomial development, for $$b>0$$

$$(1+b)^n>1+nb$$ so that

$$\frac1{(1+b)^n}<\frac1{1+nb}.$$

Then with $$a:=\dfrac1{1+b}$$,

$$a^n<\frac1{1+n\left(\dfrac1a-1\right)}=\frac{a}{a+n(1-a)}<\frac a{1-a}\frac1n\to0.$$

$$a \in (0,1)$$;

Set $$a=e^{-b}$$, where $$b >0$$.

Then

$$0\le a^x = e^{-bx}= \dfrac{1}{e^{bx}}=$$

$$\dfrac{1}{1+(bx) +(bx)^2/2!+......} <$$

$$\dfrac{1}{bx}=(1/b)\dfrac{1}{x}.$$

Take the limit $$x \rightarrow \infty$$.

You have to do nothing special, if you have already proved that, for $$0, $$\lim_{x\to\infty}a^x=0$$ Since you know logarithms, as shown in the proof of this result, you also know the property that $$\left(\frac{1}{a}\right)^{\!x}=\frac{1}{a^x} \tag{*}$$ So, let $$a>1$$ and take $$K>0$$. We want to show that, for $$x>x_0$$ (for a suitable $$x_0$$ depending on $$K$$), $$a^x>K$$. This will prove that no real number can be the limit as $$x\to\infty$$ of $$a^x$$. Why? Because if the limit is $$l$$, there is $$x_0$$ such that, for $$x>x_0$$, $$l-1. Take $$K=\max(l+1,1)$$ and you have a contradiction.

Since $$a>1$$, we have $$1/a<1$$; hence there is $$x_0$$ such that, for $$x>x_0$$, $$\left(\frac{1}{a}\right)^{\!x}<\frac{1}{K}$$ because $$\lim_{x\to\infty}(a^{-1})^x=0$$. Therefore, for $$x>x_0$$, owing to (*), $$a^x>K$$

Actually this is a proof that if $$f(x)>0$$ for $$x>x_0$$ and $$\lim_{x\to\infty}f(x)=0$$, then $$\lim_{x\to\infty}\frac{1}{f(x)}=\infty$$.