limits of polynomials A few polynomial limits:
$$\lim_{x\rightarrow 0} \frac{x^2 + 5x}{x^3+6x^2+3x}=\lim_{x\rightarrow 0} \frac{x + 5}{x^2+6x+3}=\frac{5}{3}$$
$$\lim_{x\rightarrow 0} \frac{x^4 + 5x}{x^3+6x^2+3x}=\lim_{x\rightarrow 0} \frac{x^3 + 5}{x^2+6x+3}=\frac{5}{3}$$
$$\lim_{x\rightarrow 0} \frac{x^4}{x^6}=\infty$$
Please verificate if I'm doing it correctly. For which two polynomials $W$, $P$ limit of  $W(x)/P(x)$ as x goes to $0$ can be a real number?
EDIT
Why for example 
$$\lim_{x\rightarrow0}\frac{x+1}{x^2}=\infty$$
$$\lim_{x\rightarrow0}\frac{x}{x^2+1}=0$$
 A: I am writing some facts about the question when you want to evaluate the limit of a ratinoal function with $x\to\pm\infty$ not $x\to 0$. Let $p(x)$ is a polynomial of degree $n$: $$p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$ and you want to find its limit when $x\to\pm\infty$. Then you just need to verify $p(x)$'s limit by  taking the following limit instead: $$\lim_{x\to\pm\infty}p(x)\approx\lim_{x\to\pm\infty}a_nx^n$$ For example:
$$\lim_{x\to +\infty}(-3x^4+5x^2-\pi)\sim\lim_{x\to +\infty}(-3x^4)=-3(+\infty)^4=+\infty$$ or $$\lim_{x\to -\infty}(x^9+\frac{-6}5x^2-7)\sim\lim_{x\to -\infty}(x^9)=(-\infty)^9=-\infty$$ Now for evaluating the limit of rational functions as you noted in the Edit, you can do the limit of numerator and denominator of it separately. For example: $$\lim_{x\to +\infty} \frac{x^4 + 5x}{x^3+6x^2+3x}=\frac{\lim_{x\to +\infty}(x^4 + 5x)}{\lim_{x\to +\infty}(x^3+6x^2+3x)}\sim \frac{\lim_{x\to +\infty}(x^4 )}{\lim_{x\to +\infty}(x^3)}=\lim_{x\to+\infty}x=+\infty$$
I   hope these points work for you someday during learning Calculus.
A: My thoughts: 
Let define $\lim_{x\rightarrow 0}\frac{P(x)}{Q(x)}=g$
If $P(x)=x^k\wedge k\in\mathbb{R}\implies$ limes of the nominator is equal to $0$ in all cases that it's degree is bigger than degree of $Q(x)$. Else is equal to 1
If $Q(x)=x^l\wedge l\in\mathbb{R}\implies$ limes of the denominator is equal to $0$ in all cases that it's degree is bigger than degree of $P(x)$ Else is equal to 1
In the other case, limit of $P(x)$ is equal the value of factor near the lowest power of x, of $Q(x)$ analogously.
So from this, answer is following easily.
A: Lets denote by $n$ the multiplicity of $0$ as a root of $W$ and by $k$ the multiplicity of $0$ as a root of $P$.Also denote $w(x)=\dfrac{W(x)}{x^n}$ and $p(x)=\dfrac{P(x)}{x^k}$. If $k=0$ then $\lim_{x\to0}\dfrac{W(x)}{P(x)}=\dfrac{W(0)}{P(0)}$.
If $n=k>0$ then $\lim_{x\to0}\dfrac{W(x)}{P(x)}=\dfrac{w(0)}{p(0)}$.
If $n>k>0$ then $\lim_{x\to0}\dfrac{W(x)}{P(x)}=0$.
If $n<k>0$ and $k-n$ is even then  $\lim_{x\to0}\dfrac{W(x)}{P(x)}=\text{sign}(w(0))\text{sign}(p(0))\infty$.
If $n<k>0$ and $k-n$ is odd then the $\lim_{x\to0}\dfrac{W(x)}{P(x)}$ does not exist.
