Stuck on Epsilon proof.. Using the $\epsilon-M $ definition of the limit, calculate 
$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}.$$
Working so far: 
$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}=3$$
Given $\epsilon>0$, I want M s.t. $x>M \implies \left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$
$$\left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$$
$$\left|\frac{3x^2+7-3(x^2+x+8)}{x^2+x+8} \right|<\epsilon$$
$$\left|\frac{-3x-17}{x^2+x+8} \right|<\epsilon$$
And now I'm stuck.. Any help would be great, thanks.

use the fact that $ \left | \frac 1 x - 0\right| < \delta $
$$\left| \frac{-3x - 17}{x^2 + x + 8}\right| < \left| \frac{-3x - 17}{x^2 }\right| \le  3\left |\frac 1 x \right | + 17\left |\frac 1 {x^2} \right | < 20 \delta   = \epsilon $$
