
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 $\displaystyle \lim_{x\rightarrow \infty }\frac{x+3}{x^23}=0$

1 Attachment(s)
I'm assuming your function is (x+3)/(x^29)
See attachment

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

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)/(x3)(x+3) < 2x/(x+3)(x3 since for x > 3 x + x > x+3
2x/(x+3)(x3)<  2x/(x)(x3) smaller denominator bigger fraction x < x+3
 2x/(x)(x3) = 2/(x3) x > 3
2/(x3) < e
1/(x3) < e/2
again if x 3 > e/2
1/(x3) < e/2
M = 3 + e/2