# Problem of the week #10 - June 4th, 2012

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#### Jameson

Staff member
Thank your to Chris L T521 for submitting this week's high school level problem!

$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ and $q(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots+b_1x+b_0$. Show that
$\lim_{x\to\infty}\frac{p(x)}{q(x)}=\begin{cases} \infty & \text{ if n>m}\\ \frac{a_n}{b_m} & \text{ if n=m}\\ 0 & \text{ if n<m}\end{cases}$

#### Jameson

Staff member
Congratulations to the following members for their correct solutions:

1) Sudharaka
2) Siron

Honorable mention to veronica1999 for a good intuitive explanation but not quite formal enough to constitute a proof.

Solution (from Sudharaka):

$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\mbox{ and }q(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots+b_1x+b_0$

$\Rightarrow\frac{p(x)}{q(x)}=\frac{x^n}{x^m}\left(\frac{a_n+a_{n-1}x^{-1}+\cdots+a_1x^{-n+1}+a_0x^{-n}}{b_m+b_{m-1}x^{-1}+\cdots+b_1x^{-m+1}+b_0x^{-m}}\right)$

$$\mbox{Note that, }\displaystyle\lim_{x\rightarrow\infty}\left(\frac{a_n+a_{n-1}x^{-1}+\cdots+a_1x^{-n+1}+a_0x^{-n}}{b_m+b_{m-1}x^{-1}+\cdots+b_1x^{-m+1}+b_0x^{-m}}\right)=\frac{a_n}{b_n}$$

$\therefore\lim_{x\rightarrow\infty}\frac{p(x)}{q(x)}=\frac{a_n}{b_n}\lim_{x\rightarrow\infty}x^{n-m}$

$$\mbox{Note that, }\displaystyle\lim_{x\rightarrow\infty}x^{n-m}=\begin{cases} \infty & \text{ if n>m}\\ 1 & \text{ if n=m}\\ 0 & \text{ if n<m}\end{cases}$$

$\therefore\lim_{x\rightarrow\infty}\frac{p(x)}{q(x)}=\begin{cases} \infty & \text{ if n>m}\\ \frac{a_n}{b_m} & \text{ if n=m}\\ 0 & \text{ if n<m}\end{cases}$

Q.E.D

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