If $a>1$, prove $\displaystyle{\lim_{x\to +\infty}a^x = +\infty}$ and $\displaystyle{\lim_{x\to -\infty} a^x = 0}$

Question

so for this question some of the ideas I have is to just let $\epsilon > 0$ and for $x>N$ and now I have to find an N.

My starting point is to somehow prove that $|f(x)-f(c)| < \epsilon$ so then I wrote $|x-c|< \delta$, but I'm stuck as on how to continue to actually start getting a function in terms of epsilon.

Answer 1

$a>1$, so $a^x$ and $\ln a$ are positive. For arbitrary $M,\epsilon>0$:

$$a^x>M\stackrel{a>1}\iff x>\frac{\ln M}{\ln a}$$

$$a^x<\epsilon\stackrel{a>1}\iff x<\frac{\ln \epsilon}{\ln a}$$

Answer 2

For another way,

Say $a \gt 1 $ then $a = (1 + t)$ for $t \gt 0$. And let $n \lt x$ be a natural number. Now we make use of the Binomial Theorem,

$$a^x \gt a^n = (1 + t)^n = 1 + nt + ... \gt nt $$

So if you pick $n$ large enough then you can make $a^x$ large enough which results in the first limit.

As for the second one, $$|a^{-x} - 0 | = \dfrac{1}{a^x} \lt \dfrac{1}{(1+t)^n} \lt \dfrac{1}{nt}$$

Again picking $N$ to be a natrual number greater than $\dfrac{1}{t\epsilon}$ does the trick.