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Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials

I need to prove these statements:

Let $f(x)=\sum_{j=0}^{n}a_jx^j$ , $g(x)=\sum_{j=0}^{m}b_jx^j$.

  • $$\deg(g)\gt \deg(f) \implies \lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}=0$$
  • $$\deg(g)= \deg(f) \implies \lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}=\frac{a_n}{b_n}$$
  • $$\deg(f)\gt \deg(g) \implies \lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}=\pm \infty$$

Is there any proof that would help me in all three statements, so that my answer can be shorter? I'm pretty sure I know how to do it, but I am trying to think of a cleaver way to shorten my answer.

Thanks!

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marked as duplicate by lhf, Martin Sleziak, Chris Eagle, The Chaz 2.0, Asaf Karagila May 9 '12 at 14:05

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2 Answers 2

Shortest way?

I will solve you the first one, and you do the others.

So, we have $f(x)=a_nx^n+...+a_0$ and $g(x)=b_mx^m+...+b_0$. Given that $\deg(g(x))>\deg(f(x))$, it follows that $m>n$.

Now, let us take a look on that required limit:

$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{a_nx^n+...+a_0}{b_mx^m+...+b_0}$$

Dividing the numerator and denominator by $x^m$:

$$\lim_{x\to\infty}\frac{\frac{a_nx^n}{x^m}+...+\frac{a_0}{x^m}}{\frac{b_mx^m}{x^m}+...+\frac{b_0}{x^m}}=\frac{0+...+0}{b_m+...+0}=\frac{0}{b_m}=0$$

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Suppose that $$f(x)=a_n\cdot x^n+a_{n-1}\cdot x^{n-1}+...+a_0$$ and also $$g(x)=b_m\cdot x^m+b_{m-1}\cdot x^{m-1}+...+b_0$$

Now, there are three cases:

  1. $n=m$

In this case, simply divide both polynomials by $x_n$ and $x_m$,and you get the limit as $\frac{a_n}{b_m}$.

  1. $n<m$

If you repeat the same procedure, you will see that the denominator is greater than the numerator, so you get 0.

  1. $n>m$

This is the reverse situation, numerator greater than the denominator, so it tends to infinity, but here, the sign depends on both values.

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You need to use {} for more than one character in subscript/superscript. –  The Chaz 2.0 May 9 '12 at 11:04
    
aa thanks,i will consider it –  dato datuashvili May 9 '12 at 11:07
    
Use \cdot for multiplication: $\cdot$ instead of $*$. –  Salech Alhasov May 9 '12 at 12:00

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