Find Delta for given limit of function I have : $$\lim_{x \to-2}\frac{3x+6}{x^3+8} = \frac{1}{4}$$
This is what i've come up with so far :
$$|f(x) - L| = \frac{3x+6}{x^3+8} -\frac{1}{4}| =|\frac{12x+24}{4x^3+32} -\frac{x^3+8}{4x^3+32}| = |\frac{-x^3+12x+16}{4x^3+32}|$$
I am looking for a $\delta(\epsilon)$ so that $|x - x_0| < \delta \rightarrow |f(x) - L| < \epsilon$
My problem is that I don't know how to work with that fraction .
 A: $$\left|\frac{3(x+2)}{x^3+8}-{1\over4}\right|<\epsilon\Longleftrightarrow\left|\frac{3(x+2)}{(x+2)(x^2-2x+4)}-{1\over4}\right|<\epsilon$$
$$x\ne-2\hspace{1cm}\left|\frac{3}{x^2-2x+4}-{1\over4}\right|<\epsilon$$ $$\left|\frac{-x^2+2x+8}{4(x^2-2x+4)}\right|<\epsilon\Longleftrightarrow-\epsilon<\frac{-x^2+2x+8}{4(x^2-2x+4)}<\epsilon\Longleftrightarrow \begin{cases}{x^2(-1-4\epsilon)+x(2+8\epsilon)+8-16\epsilon\over4(x^2-2x+4)}<0\\{x^2(-1+4\epsilon)+x(2-8\epsilon)+8+16\epsilon\over4(x^2-2x+4)}>0\end{cases}$$
but $4(x^2-2x+4)>0\ \forall x\in\mathbb R$, so:
\begin{cases}x^2(-1-4\epsilon)+x(2+8\epsilon)+8-16\epsilon<0\\x^2(-1+4\epsilon)+x(2-8\epsilon)+8+16\epsilon>0\end{cases}
If you solve this system, you get:
$${-1+4\epsilon+3\sqrt{1-{8\epsilon(1+2\epsilon)\over3}}\over-1+4\epsilon}<x<{-1-4\epsilon+3\sqrt{1+{8\epsilon(1-2\epsilon)\over3}}\over-1-4\epsilon}$$
that is a neighborhood of $-2$, hence:
$$2+{-1+4\epsilon+3\sqrt{1-{8\epsilon(1+2\epsilon)\over3}}\over-1+4\epsilon}<x+2<2+{-1-4\epsilon+3\sqrt{1+{8\epsilon(1-2\epsilon)\over3}}\over-1-4\epsilon}$$
$${-3+12\epsilon+3\sqrt{1-{8\epsilon(1+2\epsilon)\over3}}\over-1+4\epsilon}<x+2<{-3-12\epsilon+3\sqrt{1+{8\epsilon(1-2\epsilon)\over3}}\over-1-4\epsilon}$$
$$\left|x+2\right|<\min\left\{\left|{-3+12\epsilon+3\sqrt{1-{8\epsilon(1+2\epsilon)\over3}}\over-1+4\epsilon}\right|;{-3-12\epsilon+3\sqrt{1+{8\epsilon(1-2\epsilon)\over3}}\over-1-4\epsilon}\right\}$$
A: No when $x=-2$ the limit is $1/4$. Both the numerator and the denominator are zero. But to answer the question, the numerator of the fraction at the end of your equalities is zero at x=-2, of course. So it factors into
$$|x+2|\left|\frac{x-4}{4(x^2-2x+4}\right|.$$
Now all you have to is arrange $\delta$ to bound the second factor here. But actually, the quadratic in the denominator has minumum $3$ at $x=1$, so 
$$|x+2|\left|\frac{x-4}{4(x^2-2x+4}\right|\le \frac13|x+2||x-4|.$$
Now if you select $\delta<1$ then $|x+2|<\delta$ implies $-1<x+2<1$ from which $-7<x-4<-5$ and then $|x-4|<7$. So take $\delta=\min\{1,3\epsilon/7\}$.
Its always the same. In $|f(x)-L|$ you find a factor of $x-a$, or some other reason for it to be small when $x$ is near $a$. Then you find any bound at all on the effect of the remaining parts.
