Proving limit through definition Prove 
$$\lim_{x\to 2}\frac{x^2+4}{x+2}=2$$
through definition.
My solution:
Fix $\epsilon >0$ and find $\delta$
\begin{align}
0<|x-2|<\delta &\Rightarrow \left| \frac{x^2+4}{x+2}-2 \right| < \epsilon\\
&\Rightarrow\left|\frac{x(x-2)}{x+2}\right| < \epsilon
\end{align}
Let $\delta <1$, then $0<|x-2|<1$ then $x\in (1,3)$ and $x>0$ and $x+2>0$
$$0<|x-2|<1 \Rightarrow \frac{x}{x+2}|x-2|<\epsilon$$
In conclusion $\delta :=\min \left\{ 1,\frac{5\epsilon}3\right\}$
Unfortunately the answer is not correct (according to my book). It says $\delta :=\min \left\{ 1,\epsilon \right\}$. Where did I make a mistake?

I believe you got it wrong in the end. Remember that $1 < |x| < 3$ if we suppose $\delta < 1$. So: $$\begin{align}
0<|x-2|<\delta &\Rightarrow \left| \frac{x^2+4}{x+2}-2 \right| < \epsilon\\
&\Rightarrow\left|\frac{x(x-2)}{x+2}\right| < \frac{3\delta}{|3| - 2} = 3 \delta < \epsilon
\end{align}$$
So take $\delta = \min\{1, \epsilon/3\}$. But $\epsilon/3 < \epsilon$, so we're good.
