# Epsilon-proof as x->-infinity currently defeating me

• Dec 23rd 2010, 04:31 PM
Grep
Epsilon-proof as x->-infinity currently defeating me
I think Epsilon-delta proofs and I should spend some quality time together.

The problem is I need to prove that:

$\displaystyle \lim_{x \to -\infty} a^x = 0$

So, I need to show that, for every $\displaystyle \epsilon > 0$, there is a $\displaystyle \delta$ such that $\displaystyle |f(x) - L| < \epsilon$ whenever $\displaystyle x < \delta$.

So, given $\displaystyle \epsilon > 0$, I need to find $\displaystyle \delta$ such that:

$\displaystyle |a^x - 0| < \epsilon$ whenever $\displaystyle x < \delta$

And here's where I am stuck. I think I'll have to take a log in there somewhere. Wish my textbook wasn't so scarce on examples (Stewart's Calculus 4th edition). I sort of get limits where x->c, but I'm totally lost on the ones with infinities, and the differences between handling + and - infinities. Can't find any examples online of ones as x-> -infinity, which surely might help. Nor are the two videos at Khan Academy useful in this case.

I bet you guys get a lot of these. Surprisingly confusing for what seems a relatively simple concept, at first. Help, hints, pointers, nudges and taunts all appreciated.
• Dec 23rd 2010, 04:40 PM
Ackbeet
If you think of $\displaystyle \delta$ as a large negative number, and getting more and more negative, you'll be on the right track; also, you have to assume that $\displaystyle a>0,$ right? Otherwise, you have complex numbers floating around.

How can you simplify $\displaystyle |a^{x}-0|<\epsilon?$

If you can find a $\displaystyle \delta=\delta(\epsilon)$ that works, you'll be done. How could you do that?
• Dec 23rd 2010, 05:18 PM
Grep
Quote:

Originally Posted by Ackbeet
If you think of $\displaystyle \delta$ as a large negative number, and getting more and more negative, you'll be on the right track; also, you have to assume that $\displaystyle a>0,$ right? Otherwise, you have complex numbers floating around.

How can you simplify $\displaystyle |a^{x}-0|<\epsilon?$

Well, first and obviously, $\displaystyle |a^{x} - 0| < \epsilon \Rightarrow |a^{x}| < \epsilon$.

Then, I can assume that $\displaystyle a^{x} > 0$ so I have just:

$\displaystyle a^{x} < \epsilon$ whenever $\displaystyle x < \epsilon$.

Quote:

Originally Posted by Ackbeet
If you can find a $\displaystyle \delta=\delta(\epsilon)$ that works, you'll be done. How could you do that?

We want:

$\displaystyle a^{x} < \epsilon$ whenever $\displaystyle x < \epsilon$

Or:

$\displaystyle log_{a}(a^{x}) = log_{a}(\epsilon) \Rightarrow x = log_{a}(\epsilon)$

So I should try $\displaystyle \delta = log_{a}(\epsilon)$.

Given $\displaystyle \epsilon > 0$, we choose $\displaystyle \delta = log_{a}(\epsilon)$. Let $\displaystyle x < \epsilon$. Then

$\displaystyle |a^{x} - 0| = a^{x} < a^{\delta} = a^{log_{a}(\epsilon)} = \epsilon$

Thus

$\displaystyle |a^{x} - 0| < \epsilon$

Is that right? Egads, I think I just did it. Hey, I think I smell burnt toast! (Wink)
• Dec 23rd 2010, 07:32 PM
chisigma
Quote:

Originally Posted by Grep
I think Epsilon-delta proofs and I should spend some quality time together.

The problem is I need to prove that:

$\displaystyle \lim_{x \to -\infty} a^x = 0$

...

... that's true if $\displaystyle a >1$... of course...

http://digilander.libero.it/luposaba...ato&#91;1].jpg

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$
• Dec 23rd 2010, 07:45 PM
Grep
Quote:

Originally Posted by chisigma
... that's true if $\displaystyle a >1$... of course...

Good point, the problem description says "Let a > 1", which I should have stated. Apologies for leaving out an important part of the problem description.