Question: **Prove that $$\lim\_{x \rightarrow \infty} \frac{x^2+3}{4x^2-4x+8} = \frac{1}{4}$$ using the $\epsilon$-$\delta $ definition of a limit.** 
So we want to find a $\delta$ such that for every $\epsilon>0$ $$x > \delta \rightarrow \left|\frac{x^2+3}{4x^2-4x+8} - \frac{1}{4}\right| = \left|\frac{x+1}{4x^2-4x+8}\right| < \epsilon. $$ The ratio has a global maximum at $x=2$ Thus, $x+1 < 2x$ and $4x^2-4x+8 > 2x^2 + 8 > 2x^2$. This implies that with $ \delta >2$
$$ x > \delta \rightarrow |\frac{x+1}{4x^2-4x+8}| < \frac{2x}{2x^2} = \frac{1}{x}. $$
I think this proves it... How do I wrap it up?

Answer: 
The proof can be streamlined: if $\,p\_i\,$ are nonassociate primes then [their $\rm\color{#c00}{lcm}$ = product](https://math.stackexchange.com/a/3841440/242) so

$$ x^k\!\equiv 0\!\!\!\!\pmod{\!n}\,\Rightarrow\,n\!=\!p\_1^{e\_1}\cdots p\_n^{e\_n}\!\mid\! x^k \Rightarrow\, \forall i\!:\ p\_i\!\mid x^k\,\overset{p\rm \ prime}\Rightarrow \forall i\!: p\_i\!\mid\! x\overset{\rm\color{#c00}{lcm}}\Rightarrow p\_1\cdots p\_n\!\mid\! x$$

Conversely, for $\,e > $ all $e\_i,\,$ $\ p\_1\!\cdots p\_n\mid x\,\Rightarrow\, p\_1^{e\_1}\cdots p\_n^{e\_n}\mid (p\_1\cdots p\_n)^e\mid x^e\Rightarrow\,x^e\equiv 0\pmod{\!n}$
You should be a little more careful about the distinction between an element of $\mathbb Z\_n$ and an integer in $\mathbb Z$ that represents an element of $\mathbb Z\_n$. You have $x^k = n \ (= 0)$ in $\mathbb Z\_n$ but as integers this just means $x^k$ is congruent to $0$ modulo $n$, so $x^k = cn$ for some $c \in \mathbb Z$. You don't know that $x^k$ equals $n$ identically. Consequently, you also don't know that $p\_1, \ldots, p\_r$ are the only prime divisors of $x$, in fact throughout your proof you seem to always assume that these are the only primes that can divide any integer. This is a mistake, other prime numbers can divide integers that represent nonzero elements of $\mathbb Z\_n$.

To get $x \in (p\_1\cdots p\_r)$ you have to show that as integers $p\_1\cdots p\_r \ | \ x$. Do this a single prime at a time instead of trying to work with the prime decomposition of $x$, so show that for $1 \leq i \leq r$ you get $p\_i \ | \ x$. You can use the fact that $p\_i \ | \ n$ and $x^k = cn$ to get this.