<s>[INST] Finding limit of a quotient I can't find this for some reason. I know I asked about 6 of these before, and I was able to finish my homework but now I went back to review and I can't do a single one of these problems on my own. Even the ones I did figure out on my own. I spent probably a total of 14 hours on the homework, not sure what is wrong with me but I don't know how to factor or do basic algebra. 
Anyways I need to find 
$$\lim_{x\to -2}\frac{x+2}{x^3+8}.$$
I have spent at least an hour on it and I can't figure it out.
 [/INST] For $x=-2$, 
$$\begin{eqnarray*}
x+2 &=&0 \\
x^{3}+8 &=&0,
\end{eqnarray*}$$
which means $x=-2$ is a root of both equations. Hence $x^{3}+8$ may be
factored$^1$ as 
$$x^{3}+8=(x-(-2))Q(x)=(x+2)Q(x).\qquad(*)$$
To compute $Q(x)$


*

*you may perform the long division $(x^{3}+8):(x+2)$ (or apply the so called Riffini's rule or sinthetic division) and
find 
$$Q(x)=x^{2}-2x+4.$$

*Alternatively you write $x^{3}+8$ as
$$x^{3}+8=(x+2)(ax^{2}+bx+c),$$
(you could have made $a=1$, because the coefficient of $x^3$ is $1$) expand the RHS 
$$(x+2)(ax^{2}+bx+c)=ax^{3}+bx^{2}+cx+2ax^{2}+2bx+2c,$$
group the terms of the same degree 
$$(x+2)(ax^{2}+bx+c)=ax^{3}+\left( b+2a\right) x^{2}+\left( c+2b\right) x+2c$$
and equate to $x^{3}+8$ 
$$ax^{3}+\left( b+2a\right) x^{2}+\left( c+2b\right) x+2c=x^{3}+8.$$
The two sides are the same polynomial if and only if their coefficients are
equal
$$\begin{eqnarray*}
a &=&1 \\
b+2a &=&0 \\
c+2b &=&0 \\
2c &=&8,
\end{eqnarray*}$$
which means $a=1,b=-2,c=4$. 
So
$$x^{3}+8=(x+2)(x^{2}-2x+4).\qquad(**)$$
Thus, for $x\neq -2$, we have
$$
\frac{x+2}{x^{3}+8}=\frac{x+2}{\left( x+2\right) \left( x^{2}-2x+4\right) }=
\frac{1}{x^{2}-2x+4},\qquad(***)$$
i.e $\frac{x+2}{x^{3}+8}$ is equal to $\frac{1}{x^{2}-2x+4}$, except for $x=-2
$. Now you can safely compute the limit 
$$\lim_{x\rightarrow -2}\frac{x+2}{x^{3}+8}=\lim_{x\rightarrow -2}\frac{1}{
x^{2}-2x+4}=\dfrac{1}{\displaystyle\lim_{x\rightarrow -2}
x^{2}-2x+4}=\dfrac{1}{(-2)^{2}-2(-2)+4}=\dfrac{1}{12}.$$ 
--
$^1$ If a polynomial of degree $n$, 
$$P(x)=a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\ldots +a_{n-1}x+a_{n}\quad
(a_{0}\neq 0),$$
has as roots $n$ different  numbers $\alpha _{1},\alpha _{2},\ldots
,\alpha _{n}$ , then it may be factored as
$$P(x)=a_{0}(x-\alpha _{1})(x-\alpha _{2})\cdots (x-\alpha _{n}).$$
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