# Epsilon-delta limit proof

• Jul 31st 2010, 09:30 PM
Glitch
Epsilon-delta limit proof
The question:

Prove that the limit as x goes to infinity is 0 for $\frac{x+3}{x^2-3}$

My attempt:

$|\frac{x+3}{x^2-3} - 0| < \varepsilon$

Now I'm stuck. How do I solve this for x? I'm sure this is simple, but for some reason I'm just not seeing it. :/

Thanks.
• Jul 31st 2010, 10:10 PM
TheEmptySet
Quote:

Originally Posted by Glitch
The question:

Prove that the limit as x goes to infinity is 0 for $\frac{x+3}{x^2-3}$

My attempt:

$|\frac{x+3}{x^2-3} - 0| < \varepsilon$

Now I'm stuck. How do I solve this for x? I'm sure this is simple, but for some reason I'm just not seeing it. :/

Thanks.

To show a limit at infinity we need to show that for every $\epsilon > 0$ there exists an $N \in \mathbb{R}$ such that for $x > N$ $|f(x)-L|< \epsilon$

Notice that for n > 3

$\displaystyle \bigg|\frac{N+3}{N^2-3} \bigg |< \bigg| \frac{N+3}{N^2-9}\bigg| =\bigg| \frac{N+3}{(N+3)(N-3)}\bigg|=\bigg| \frac{1}{N-3}\bigg|$
• Jul 31st 2010, 10:36 PM
Glitch
I'm having trouble following. I think I understand the general proof and conditions, but I'm unsure of why you substituted N for x.
• Jul 31st 2010, 10:49 PM
TheEmptySet
To show that the limit exists we need to show that when x is large enough that the limit is close to 0. So I picked N to be my large x. So now for any x > N the inequality will hold.