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?

Given that $x \in (1,3)$, we want to find an upper bound for:
$$
\frac{x}{x+2}
$$
To increase the value of this fraction, we can maximize the numerator and minimize the denominator. Indeed, since $x < 3$ and $x + 2 > 3$, we have that:
$$
\frac{x}{x+2} < \frac{3}{3} = 1
$$
Thus, we have that:
$$
\frac{x}{x+2}|x-2| < |x -2| < \epsilon
$$
provided that we chose $\delta :=\min \left\{ 1,\epsilon \right\}$.
