Prove by limit definition $\lim _{x\to \infty }\left(\frac{-7x^2+9x}{4x^2+8}\right)=\frac{-7}{4}$ Prove by limit definition
$$\lim _{x\to \infty }\left(\frac{-7x^2+9x}{4x^2+8}\right)=\frac{-7}{4}$$
let $\epsilon > 0$   need to find $M$ such that for every $x>M \implies |f(x) - L|<\epsilon$ 
$\left|\frac{-7x^2+9x}{4x^2+8}+\frac{7}{4}\right| = \frac{9x+14}{4\left(x^2+2\right)} \le \frac{23x}{4x^2}=\frac{23}{4x} < \epsilon $
so I choose $M=\frac{23}{4\epsilon}$
My question is that I assumed that $x>0$ do I need to check when $x\le0$ or this is enough since  $x \to \infty $ ? and does this prove the limit ? 
thanks 

Look for example at pp. 105, Calculus (Third Edition) from Spivak. The definition of $\lim_{x\to\infty} f(x)=L$ is that for every $\varepsilon>0$ there is a number $N$ such that, for all $x$, if $x>N$, then $|f(x)-L|<\varepsilon$. According to this, what you have done proves the limit, but I think that in your attempt to solve the problem it was necessary to assume $x>1$. Nevertheless this is not a problem because you can take $M=\max\{1, \frac{23}{4\varepsilon}\}$ to ensure that the bounds you took are ok.
