Delta Epsilon Proof for Ratio of Polynomials 
Prove that $f$ is continuous at $x_0$ using the $\delta$ and $\epsilon$ criteria for:
$f(x)=\frac{x^3-2}{x+3}$, $x_0=1$.

I've made it halfway there, about; since $$|\frac{x^3-2}{x+3}+\frac{1}{4}|=|x-1||\frac{4x^2+4x+5}{4(x+3)}|$$
So then the issue is that we need $\delta>|x-1|$ but also bigger than the other expression.  The only issue is, I don't really know how to find such a $\delta$.  How can you proceed from here?
 A: Since you want $\;x\;$ "close" to $\;x_0=1\;$ , you can decide that $\;\frac12<x<2\;$ say,  so (assuming your work is correct):
$$|x-1|\left|\frac{4x^2+4x+5}{4(x+3)}\right|\le|x-1|\frac{29}{14}$$
so the right hand expression will be less than $\;\epsilon\;$ if, for example, you take 
$$\delta:=\frac{14}{29}\epsilon\implies|x-1|\left|\frac{4x^2+4x+5}{4(x+3)}\right|<\epsilon$$
and we're done.
A: $x>0$, and consider $|x-1| \lt 1$, i.e. $0<x<2$.
Let $\epsilon >0$ be given.
Choose $\delta < \min(1,\dfrac {12}{29}\epsilon)$
Then:
$|x-1|\dfrac{4x^2+4x+5}{4(x+3)} \lt$
$|x-1|\dfrac{4(2^2) +4(2) +5}{4(3)} =$
$|x-1| \dfrac{29}{12} \lt \delta \dfrac{29}{12} \lt \epsilon$.
A: Suppose that $\delta <1$ i.e. $0<x<2$ and bound rational function that is giving you trouble on this interval with a bound say $M$. Then take $\delta$ to be the minimum of $1$ and $\varepsilon /M$.
A: This is a bit long but I wanted to go into detail so that you could apply this approach more easily to other examples.
Unless the function has a vertical tangent at $x_0$ there will be some positive bound $B$ satisfying
\begin{equation}
\left\vert \frac{f(x)-L}{x-x_0}\right\vert\le B \tag{1}
\end{equation}
on some deleted $c$-neighborhood $(x_0-c,x_0)\cup(x_0,x_0+c)$. Then if we let
\begin{equation}
\delta=\min\left\{c,\frac{\epsilon}{B}\right\} \tag{2}
\end{equation}
it follows that
\begin{eqnarray}
\left\vert \frac{f(x)-L}{x-x_0}\right\vert\cdot\left\vert x-x_0 \right\vert<B\cdot\frac{\epsilon}{B}\\
\left\vert f(x)-L\right\vert<\epsilon
\end{eqnarray}
For this particular exercise we wish to find a $B$ satisfying
\begin{equation}
\left\vert\frac{\frac{x^3-2}{x+3}+\frac{1}{4}}{x-1} \right\vert\le B
\end{equation}
For $x\ne1$ this simplifies to
\begin{equation}
\left\vert (x-2)+\frac{29}{4(x+3)}\right\vert\le B \tag{3}
\end{equation}
So let us investigate whether there is such a bound $B$ if we choose $c=1$ from equation (2) above.
\begin{equation}
\vert x-1\vert<1
\end{equation}
Then begin with 
\begin{equation}
-1<x-1<1
\end{equation}
and try to work our way to an expression that looks like inequality $(3)$.
\begin{eqnarray}
-1 &<x-1<1\\
-2 &<x-2<0
\end{eqnarray}
Now we must add in the $\frac{29}{4(x+3)}$ term. We know its graph is an hyperbola which is decreasing to the right of its vertical asymptote at $x=-3$ so its largest value on $(0,1)
\cup(1,2)$ is $\frac{29}{12}$ at $x=0$ and its smallest value is $\frac{29}{20}$ at $x=2$. Therefore we continue
\begin{array}
 \phantom{ }-2&<&x-2&<0\\
\tfrac{29}{20}&<&\frac{29}{4(x+3)}&<\frac{29}{12}\\
-2+\tfrac{29}{20}&<&x-2+\frac{29}{4(x+3)}&<\frac{29}{12}\\
-\tfrac{11}{20}&<&x-2+\frac{29}{x+3}&<\frac{29}{12}\\
&&\left\vert4(x-2)+\frac{29}{4(x+3)} \right\vert&<\frac{29}{12}<3
\end{array}
So let $\delta=\min\left\{1,\frac{\epsilon}{3}\right\}$. Then $\vert x-1\vert<1$ and $\vert x-1\vert<\frac{\epsilon}{3}$. Therefore
\begin{eqnarray}
\left\vert(x-2)+\frac{29}{4(x+3)} \right\vert\cdot\vert x-1\vert&<3\cdot\frac{\epsilon}{3}\\
\left\vert\frac{x^3-2}{x+3}+\frac{1}{4} \right\vert&<\epsilon
\end{eqnarray}
Therefore $\lim_{x\to1}f(x)=-\frac{1}{4}=f(1)$ so the function is continuous at $x=1$.
