Proving that $\lim_{x\to 2}\frac{x^2-5x}{x^2+2}=-1$ using the $\epsilon$-$\delta$ definition of a limit My attempt: 
$$
\left|\frac{x^2-5x}{x^2+2}+1\right|<\left|\frac{x^2-5x}{x^2+2}\right|<
\left|\frac{x^2-5x}{x^2}\right|<\frac{1}{x^2}|x^2-5x|,$$
using the restriction $|x-2|<2$, so $0<x<4$, thus
$$\frac{1}{x^2}|x^2-5x|=\frac{1}{x^2}|x^2-6x+4-4+x|=\frac{1}{x^2}|(x-2)^2+x-4|<\frac{(x-2)^2}{x^2}<\epsilon,$$
hence, due to the fact that $|x-2|<\delta\le2$ we get
$$
\delta^2<\epsilon x^2 < \epsilon16.
$$
Finally, for any $\epsilon >0$ the corresponding $\delta$ is $\min\{\epsilon, 4\epsilon^{1/2}\}$.
Is it correct?

Let $\epsilon>0$ be given. We have to find a number $\delta>0$ such that $\left|\frac{x^2-5x}{x^2+2}+1\right|<\epsilon$ whenever $|x-2|<\delta$. But, as Andre notes,
$$
\left|\frac{x^2-5x}{x^2+2}+1\right|=\left|\frac{2x^2-5x+2}{x^2+2}\right|=|x-2|\left|\frac{2x-1}{x^2+1}\right|.
$$
We find a positive constant $C$ such that $\left|\frac{2x-1}{x^2+1}\right|<C\Rightarrow |x-2|\left|\frac{2x-1}{x^2+1}\right|<C|x-2|$, and we can make $C|x-2|<\epsilon$ by taking $|x-2|<\frac{\epsilon}{C}=\delta$. We restrict $x$ to lie in the interval $|x-2|<1$ and note the following:
\begin{align}
|x-2|<1&\implies 1<x<3\\[1em]
&\implies 1>\frac{1}{x}>\frac{1}{3}\\[1em]
&\implies 1>\frac{1}{x^2}>\frac{1}{9}\\[1em]
&\implies \frac{1}{3}>\frac{1}{x^2+2}>\frac{1}{11}\\[1em]
&\implies \frac{5}{3}>\frac{2x-1}{x^2+2}>\frac{1}{11}\\[1em]
&\implies C=\frac{5}{3}.
\end{align}
Thus, we should choose $\delta=\min\left\{1,\frac{3\epsilon}{5}\right\}$. To see that this choice of $\delta$ works, consider the following:
Given $\epsilon>0$, we let $\delta=\min\left\{1,\frac{3\epsilon}{5}\right\}$. If $|x-2|<1$, then $\left|\frac{2x-1}{x^2+2}\right|<\frac{5}{3}$. Also, $|x-2|<\frac{3\epsilon}{5}$. Hence,
$$
\left|\frac{x^2-5x}{x^2+2}+1\right|=|x-2|\left|\frac{2x-1}{x^2+1}\right|<\frac{3\epsilon}{5}\cdot\frac{5}{3}=\epsilon,
$$
as desired. $\blacksquare$
