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?
 A: 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 $\delta :=\min \left\{ 1,\epsilon \right\}$.
A: 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.
A: Be aware that multiple $\delta$ will work for any given $\epsilon>0.$ Note, for example, that if $\delta=\delta_0$ works, then $\delta=\frac12\delta_0$ also works--and more generally, any $0<\delta<\delta_0$ works just fine.
The $\delta$ you found still does the trick. The book simply found a (potentially) smaller one.
You haven't quite justified your $\delta,$ however. Instead, you might consider the following approach.

If $0<\delta\le 1,$ then for $0<|x-2|<\delta,$ we have that $x\in(1,3),$ so in particular $x>0$ and $x+2>0,$ and so $$\left|\frac{x(x-2)}{x+2}\right|=\frac{x}{x+2}|x-2|<\frac{x}{x+2}\delta.$$ Observing moreover that $$\frac{x}{x+2}=1-\frac2{x+2}<1-\frac2{3+2}=\frac35$$ whenever $x\in(1,3),$ it follows that for such $\delta$ and for $0<|x-2|<\delta,$ we have $$\left|\frac{x(x-2)}{x+2}\right|<\frac35\delta.$$ Therefore, putting $\delta=\min\left\{1,\frac{5\epsilon}3\right\},$ we see that....

A few spots could probably use a bit more justification than that, but that should at least give you an idea.
A: I have been staring at my PC-screen in disbelief. Anyone with a background in mathematics, when given this exercise, will make the following instantaneous observations: 
The denominator of $f(x)$ is perfectly well-behaved and far from a zero in the neighbourhood of $x = 2$. The same is true for the numerator of $f(x)$. Therefore the limit for $x$ to $2$ is found by direct substitution. The limit is simply $f(2) = 2$. 
Since there are no complications whatsoever with the above reasoning, I feel that it would be excessive to ask for a proof. However, a teacher or scientist in a good mood may point out that the Taylor series of $f(x)$ around $x = 2$ is well-behaved. He may even compute the first few terms. That should suffice!
The epsilon-delta method might be rigorous, but it is also cumbersome and in cases such as this one unnecessary.
