# Limits

• Feb 3rd 2010, 12:14 PM
vuze88
Limits
Use the epsilon definition of the limit that $\displaystyle \lim_{x\rightarrow \infty }f(x)=L$ if for every positive real number $\displaystyle \epsilon$, there exists a real number $\displaystyle M$ such that $\displaystyle \left | f(x)-L\right |< \epsilon$ whenever $\displaystyle x>M$, to prove $\displaystyle \lim_{x\rightarrow \infty }\frac{x+3}{x^2-3}=0$
• Feb 3rd 2010, 06:46 PM
Calculus26
I'm assuming your function is (x+3)/(x^2-9)

See attachment
• Feb 3rd 2010, 11:04 PM
vuze88
thanks for the help. So is it not possible to do the initial question?
• Feb 4th 2010, 03:07 AM
Calculus26
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