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: 
You can do lengthy computations with elements, or avoid them.

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First the method without computations. If $R$ is a commutative ring and $I$ is an ideal of $R$, then the prime ideals of $R/I$ are of the form $P/I$, where $P$ is a prime ideal of $R$ and $I\subseteq P$. Moreover, the ideal of nilpotent elements is the intersection of all prime ideals.

Indeed, an ideal $R/I$ is of the form $J/I$ with $I\subseteq J$ and $R/J\cong (R/I)/(J/I)$, so $R/J$ is a domain if and only if $(R/J)/(J/I)$ is a domain.

Now the prime ideals of $\mathbb{Z}$ containing $n\mathbb{Z}$ are those of the form $p\mathbb{Z}$, with $p\mid n$. So, in your notation, they are $p\_1\mathbb{Z}, p\_2\mathbb{Z},\dots,p\_r\mathbb{Z}$. Thus the prime ideals of $\mathbb{Z}/n\mathbb{Z}$ are the quotients thereof, so the nilradical of $\mathbb{Z}/n\mathbb{Z}$ is
$$
\bigcap\_{1\le i\le r}\frac{p\_i\mathbb{Z}}{n\mathbb{Z}}=
\frac{p\_1p\_2\dots p\_r\mathbb{Z}}{n\mathbb{Z}}
$$
This ideal is trivial if and only if $n=p\_1p\_2\dots p\_r$.

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The method with computations. Let $[x]$ denote the element in $\mathbb{Z}/n\mathbb{Z}$ corresponding to $x\in\mathbb{Z}$.

In order that $[x]^k=[0]$ for some $k>0$, we need that $n\mid x^k$ and, in particular, that every prime dividing $n$ also divides $x^k$, hence $x$. So, if $[x]$ is nilpotent, $p\_1p\_2\dots p\_r$ divides $x$.

Conversely, if $a=\max\{a\_1,a\_2,\dots,a\_r\}$, we have that
$$
n\mid (p\_1p\_2\dots p\_k)^a
$$
because $p\_i^{a\_i}$ divides $(p\_1p\_2\dots p\_k)^a$, for $i=1,2,\dots,r$. Thus $[p\_1p\_2\dots p\_r]$ is nilpotent.

Therefore, the ideal generated by $[p\_1p\_2\dots p\_r]$ is nilpotent.

The two inclusions having been proved, we have that the nilradical is the ideal generated by $[p\_1p\_2\dots p\_k]$.

Finally, if $a>1$, the ideal generated by $[p\_1p\_2\dots p\_r]$ is a nonzero ideal of $\mathbb{Z}/n\mathbb{Z}$, so the ring $\mathbb{Z}/n\mathbb{Z}$ has no nonzero nilpotent element if and only if $n$ is squarefree.
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.