Prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ The title is quite clear. I am required to prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ using the epsilon-delta definition of the limit.

Given any $\varepsilon \gt 0$, there exists a $\delta =$
Such that $0 \lt \lvert x-1 \rvert \lt \delta \Rightarrow \lvert\frac{x+2}{x^2+1} -\frac{3}{2} \rvert \lt\varepsilon$
$\frac{x+2}{x^2+1} -\frac{3}{2}= \frac{-3x^2+2x+1}{2(x^2+1)}$
I have managed to express both the numerator and the denominator as such
$-3x^2+2x+1 = -(x-1)(3x+1)\\
2(x^2+1) = 2x^2+2= 2(x-1)^2 + 4(x-1) +4$
Returning back to the fraction $\frac{-(x-1)(3x+1)}{2(x-1)^2 + 4(x-1) +4}$

I am unsure on how to continue and would really appreciate some guidance.

So you simplified the expression 
\begin{align}
\left | \frac{x+2}{x^2+1} -\frac{3}{2} \right | = \left | \frac{(x-1)(3x+1)}{2x^2+2} \right |.
\end{align}
The good news is we have an $|x-1|$ appearing, but there is also a $|3x+1|$ and $|2x^2+2|,$ which are expressions we don't necessarily want. The thing to do here is to bound these unwanted expressions by numbers, which I demonstrate in this problem.
Suppose $\delta < 1.$ What happens if $|x-1|< \delta$  ?
First, we see $0 < x < 2.$ Multiplying all sides of the inequality by $3$ and then adding $1$ to all sides, we see $1<3x+1<7,$ so $|3x+1|<7.$ Second, we see upon squaring all sides of $0<x<2$ and then adding $2$, $2<2x^2+2<10,$ so $\frac{1}{|2x^2+2|} < \frac{1}{2}.$
Now, for any $\epsilon>0,$ let us set $\delta= \min \lbrace 1, \frac{2 \epsilon}{7} \rbrace.$ Why does this work ?
If $\delta=1,$ then we see $|x-1| < \delta$ implies  $|x-1|<\frac{2 \epsilon}{7}$ and also
\begin{align}
\left|f(x)-\frac{3}{2} \right| = \left | \frac{x+2}{x^2+1} -\frac{3}{2} \right | = \left | \frac{(x-1)(3x+1)}{2x^2+2} \right |< \frac{7|x-1|}{2}< \frac{7}{2} \frac{2 \epsilon}{7}= \epsilon
\end{align} 
If $\delta=\frac{2 \epsilon}{7}$ then already we have $|x-1|< \delta$ implies
\begin{align}
\left|f(x)-\frac{3}{2} \right|=\left | \frac{x+2}{x^2+1} -\frac{3}{2} \right | = \left | \frac{(x-1)(3x+1)}{2x^2+2} \right |< \frac{7|x-1|}{2}< \frac{7}{2} \frac{2 \epsilon}{7}= \epsilon.
\end{align} 
