Proving $\lim\limits_{x\rightarrow 1} \frac{x^2+3}{x+1}=2$ using the formal definition of the limit 
Prove $\lim\limits_{x\rightarrow 1} \frac{x^2+3}{x+1}=2$ using the formal definition of the limit.

My question is, I've picked $\delta\lt1$, and I've found that $\delta \lt \min(1,\sqrt{\epsilon})$. Was picking $1$ problematic at all? and is my choice for $\delta$ correct?
Rest of Proof:
$$0\lt|x-1|\lt \delta \Rightarrow \left|\frac{(x-1)^2}{x+1}\right|\lt \epsilon$$
Picking $\delta \lt 1$:
$$|x-1|\lt 1 \Rightarrow -1\lt x-1 \lt 1$$
And we get from that $\frac13 \lt \frac{1}{x+1} \lt 1$ which leads to $\left|\frac{1}{x+1}\right| \lt 1$
Let's go back:
$$\left|\frac{(x-1)^2}{x+1}\right|\lt \left|1(x-1)^2\right|\lt \epsilon$$ 
Since $(x-1)^2\gt 0$ we can get rid of the absolute value
and we get 
$$(x-1)^2\lt \epsilon \rightarrow x-1 \lt \sqrt{\epsilon}$$
Also: What is the difference between picking $\delta=1$ and $\delta \lt 1$

Here is a more "gimmicky" kind of solution (Battani's is definitely more elegant and natural): We find a positive constant $C$ such that $\left|\frac{x-1}{x+1}\right|<C\Rightarrow |x-1|\left|\frac{x-1}{x+1}\right|<C|x-1|$, and we can make $C|x-1|<\epsilon$ by taking $|x-1|<\frac{\epsilon}{C}=\delta$. We restrict $x$ to lie in the interval $|x-1|<1$ and note the following:
\begin{align}
|x-1|<1&\implies 0<x<2\\[1em]
&\implies 1<x+1<3\\[1em]
&\implies 1>\frac{1}{x+1}>\frac{1}{3}\\[1em]
&\implies 3>\frac{x-1}{x+1}\\[1em]
&\implies C=3.
\end{align}
Thus, we should choose $\delta=\min\left\{1,\frac{\epsilon}{3}\right\}$. To see that this choice of $\delta$ works, consider the following: Given $\epsilon>0$, we let $\delta=\min\left\{1,\frac{\epsilon}{3}\right\}$. If $|x-1|<1$, then $\left|\frac{x-1}{x+1}\right|<3$. Also, $|x-1|<\frac{\epsilon}{3}$. Hence,
$$
\left|\frac{x^2+3}{x+1}-2\right|=|x-1|\left|\frac{x-1}{x+1}\right|<\frac{\epsilon}{3}\cdot3=\epsilon,
$$
as desired. $\blacksquare$
