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 $\delta \lt 1$

It  seems me correct,the alternative short proof could be:
For any $\epsilon >0$ ,and  by choosing $\quad \delta =min\left( 1,\epsilon  \right) $ and  note that if $\left| x-1 \right| <\delta $ the we will get

$$
\left| \frac { x^{ 2 }+3 }{ x+1 } -2 \right| =\left| \frac { { x }^{ 2 }-2x+1 }{ x+1 }  \right| =\left| \frac { { \left( x-1 \right)  }^{ 2 } }{ \left( x-1 \right) +2 }  \right| <\left| \frac { { \left( x-1 \right)  }^{ 2 } }{ \left( x-1 \right)  }  \right| =\left| x-1 \right| <\epsilon 
$$

