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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

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    $\begingroup$ I think your proof is perfect :) $\endgroup$ – Mike Earnest Sep 28 '18 at 19:24
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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 $x_0<k<\sqrt{x_0}$. Hence, $k^2<x_0$.

By the property of $\sup$, we can find a $l$ such that $k<l<x_0$ 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, $$k^{2n}<l^n<\frac12$$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 $$\lim_{n\to\infty}\bigg(\frac12\bigg)^n=0$$

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By the Binomial development, for $b>0$

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

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

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

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

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$a \in (0,1)$;

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 $x \rightarrow \infty$.

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You have to do nothing special, if you have already proved that, for $0<a<1$, $$ \lim_{x\to\infty}a^x=0 $$ 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 $x>x_0$, $l-1<a^x<l+1$. 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 $\lim_{x\to\infty}\frac{1}{f(x)}=\infty$.

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