<s>[INST] 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.
 [/INST] Let $x>\max\{\sqrt{17},\frac{3}{\epsilon-1}\}$. We have $x>0$. This gives us
$$\left|\frac{-3x-17}{x^2+x+8} \right|= \left|\frac{3x+17}{x^2+x+8} \right|=\frac{3x+17}{x^2+x+8}<\frac{3x+17}{x^2}<\frac{3}{x}+1<\epsilon$$
This is incomplete as it assumes $\epsilon\neq 1$. When $\epsilon=1$ you can solve a quadratic and find which $x$'s work.
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