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.

Your inequality can be written
$$
\left|\frac{(x-1)(3x+1)}{x^2+1}\right|<2\varepsilon\tag{*}
$$
If $x>0$, then $x^2+1>1$, so $\frac{1}{x^2+1}<1$. The condition that $x>0$ is satisfied as soon as $\delta<1$, when you take $0<|x-1|<\delta$.
In this case, we have
$$
\left|\frac{(x-1)(3x+1)}{x^2+1}\right|<
\lvert(x-1)(3x+1)\rvert
$$
Since also $3x+1<7$, because $x<2$, we have
$$
\lvert(x-1)(3x+1)\rvert<7\lvert x-1\rvert<7\delta
$$
If we take $\delta=\min\{2\varepsilon/7,1\}$, we achieve what we want: for $0<|x-1|<\delta$, we have
$$
\left|\frac{(x-1)(3x+1)}{x^2+1}\right|<
\lvert(x-1)(3x+1)\rvert
<7\lvert x-1\rvert<7\delta\le7\frac{2\varepsilon}{7}=2\varepsilon
$$
The solution set of (*) may well be larger than the set described by
$$
0<\lvert x-1\rvert<\min\!\left\{\frac{2\varepsilon}{7},1\right\}
$$
but this is unimportant: what's needed for checking the correctness of a limit is that to every $\varepsilon>0$ there corresponds a $\delta>0$ et cetera: once you find a value for $\delta$ all numbers $\delta'$ with $0<\delta'<\delta$ would be good as well.
In some applications related to approximations, getting “the largest $\delta$” can be important, but not for the proof of correctness.
