# Math Help - Limits

1. ## Limits

Use the epsilon definition of the limit that $\lim_{x\rightarrow \infty }f(x)=L$ if for every positive real number $\epsilon$, there exists a real number $M$ such that $\left | f(x)-L\right |< \epsilon$ whenever $x>M$, to prove $\lim_{x\rightarrow \infty }\frac{x+3}{x^2-3}=0$

2. I'm assuming your function is (x+3)/(x^2-9)

See attachment

3. thanks for the help. So is it not possible to do the initial question?

4. No it is very possible though a little more difficult-- my impression is that at this level it would seem the one I suggested would be more relevant.

However for the original

| (x + 3)/(x-3)(x+3)| < |2x/(x+3)(x-3| since for x > 3 x + x > x+3

|2x/(x+3)(x-3)|< | 2x/(x)(x-3)| smaller denominator bigger fraction x < x+3

| 2x/(x)(x-3)| = 2/(x-3) x > 3

2/(x-3) < e

1/(x-3) < e/2

again if x- 3 > e/2

1/(x-3) < e/2

M = 3 + e/2