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.

In this answer we continue where the OP left off, making one last 'normalization'  change:
$\tag 1 \frac{-(x-1)(3x+1)}{2(x-1)^2 + 4(x-1) +4} = \frac{-(x-1)(3(x-1)+4)}{2(x-1)^2 + 4(x-1) +4}$

Denominator:
If $|x - 1| \lt 1/2$ then
$\quad |2(x-1)^2 + 4(x-1) +4| \gt  |4(x-1) +4| \gt 2 $
Note that $4(x-1) \gt -2$.

$\text{ }$

Numerator:
If $|x - 1| \lt 1/2$ then  
$\quad \;  | -(x-1)\;(3\,(x-1)+4) | = |x-1| \, |3(x-1)+4| $
$ \qquad \quad \le | x-1|\;( 3 \,|x-1| + 4) \lt 5.5 \, |x-1| $
Use the triangle inequality.

Now the absolute value of the ratio expression (1) gets bigger as the numerator gets larger and the denominator gets smaller. So if $|x - 1| \lt 1/2$ and since $\frac{5.5}{2} = \frac{11}{4}$,
$\tag 2 \text{Absolute value of (1) } \lt \frac{11}{4} |x -1|$
We are now ready! Let the $\varepsilon \gt 0$ challenge be given and set
$\qquad \delta = min(\frac{1}{2},\frac{4 \varepsilon}{11})$
