Limit of quotient of 2 polynomials Given that $a$ is a zero with multiplicity $5$ for a polynomial $p$, and a zero with multiplicity $7$ for a polynomial $q$. Determine $\lim_{x \to a} \frac{p(x)}{q(x)}$.
Alright, my take on this: $p(x) = (x - a)^5$ and $q(x) = (x - a)^7$, so: $\lim_{x \to a} \frac{p(x)}{q(x)} = \lim_{x \to a} \frac{(x - a)^5}{(x - a)^7} = \lim_{x \to a} \frac{1}{(x - a)^2}$. 
Since $x$ approaches $a$, but never quite reaches it. I think we have two choices: 


*

*$x$ approaches $a$ from the left, so it's a little bit smaller than
$a$, making $(x-a)$ a negative number. But since we have $(x - a)^2$ the result is still positive.

*$x$ approaches $a$ from the right, so it's a little bit bigger than
    $a$, making $(x-a)$ a positive number.
In either case the result is really small, dividing 1 by something really small, should result in something really big. But this isn't correct. How should I approach this? The more I think about it, the more confused/insecure I get :/
EDit: I think I get it:
$\lim_{x \to a} \frac{(x - a)^5r(x)}{(x - a)^7(s(x)} =\lim_{x \to a} \frac{r(x)}{(x - a)^2s(x)}$. Since $\lim_{x \to a}(x - a)^2s(x) \to 0$, then according to the division laws of limits, it should not exist.
 A: You cannot assume that $p(x)=(x-a)^5$. But you can write that $p(x)=(x-a)^5*r(x)$ where $r$ is a polynomial s.t. $r(a)\neq 0$. Same for $q(x)=(x-a)^7*s(x)$.
Then $\frac{p(x)}{q(x)}=\frac{r(x)}{s(x)(x-a)^2} \rightarrow_{x\rightarrow a}\infty$, the sign in front of $\infty$ being the same on both sides of $a$ and the same as the sign of $\frac{r(a)}{s(a)}$.
Given the possible answers you mention, the correct one is "The limit of $\frac{p(x)}{q(x)}$ when $x\rightarrow a$ does not exist."
A: We have:
$\lim_{x \to a} \frac{p(x)}{q(x)} = \lim_{x \to a} \frac{(x - a)^5}{(x - a)^7} = \lim_{x \to a} \frac{1}{(x - a)^2}= \infty$.
A: I'll make an answer to this (hoping its correct):
Given that $a$ is a zero with multiplicity $5$ for a polynomial $p$, and a zero with multiplicity $7$ for a polynomial $q$. Determine $\lim\limits_{x \to a} \frac{p(x)}{q(x)}$.
We have that $p(x) = (x - a)^5r(x)$ and $q(x) = (x - a)^7s(x)$. 
Now $\lim\limits_{x \to a} \frac{(x - a)^5r(x)}{(x - a)^7s(x)} = \lim\limits_{x \to a} \frac{r(x)}{(x - a)^2s(x)}$. 
Since we know $\lim\limits_{x \to a}(x - a)^2$ is $0$, the denominator is $0$.
I need to apply the following law of limits of quotients here:
If $\lim\limits_{x \to a} f(x) = L$ and $\lim\limits_{x \to a} f(x) = M$
If $M = 0$, but $L \neq 0$, then $\lim\limits_{x \to a} \frac{f(x)}{g(x)}$ does not exist.
