The question:

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

My attempt:

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

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.