Polinomials and Limit

**osodud** · June 16th 2009, 10:50 AM

Hello

Could you PLEASE help me to solve the following exercise:

Let $p(x)$ and $q(x)$ polinomials of order $n$ and $m$ respectively.

Analyze what happens when $p(x)$ / $q(x)$ and $limx->0$

THANKS IN ADVANCE

**pankaj** · June 16th 2009, 11:41 AM

$p(x)=a_{0}+a_{1}x+a_{2}x^2+.....+a_{n}x^n$

and

$q(x)=b_{0}+b_{1}x+b_{2}x^2+.....+b_{m}x^m$

$ \lim_{x\to 0}\frac{p(x)}{q(x)}=\frac{a_{0}}{b_{0}} $

Nothing special about this.
.................................................. ...............................................
Now consider
$p(x)=a_{0}x^n+a_{1}x^{n+1}+a_{2}x^{n+2}2+.....+a_{ n}x^{n+k}$

$q(x)=b_{0}x^m+b_{1}x^{m+1}+b_{2}x^{m+2}+.....+b_{n }x^{m+l}$

Evaluate $\lim_{x\to 0}\frac{p(x)}{q(x)}$ under the following conditions:

(i) $n=m$

(ii) $n>m,n-m$ is even

(iii) $n<m,n-m$ is even, $\frac{a_{0}}{b_{0}}>0$

(iv) $n<m,n-m$ is even, $\frac{a_{0}}{b_{0}}<0$

**Danny** · June 16th 2009, 11:59 AM

Originally Posted by pankaj

$p(x)=a_{0}+a_{1}x+a_{2}x^2+.....+a_{n}x^n$

and

$q(x)=b_{0}+b_{1}x+b_{2}x^2+.....+b_{m}x^m$

$ \lim_{x\to 0}\frac{p(x)}{q(x)}=\frac{a_{0}}{b_{0}} $

Nothing special about this.
.................................................. ...............................................
Now consider
$p(x)=a_{0}x^n+a_{1}x^{n+1}+a_{2}x^{n+2}2+.....+a_{ n}x^{n+k}$

$q(x)=b_{0}x^m+b_{1}x^{m+1}+b_{2}x^{m+2}+.....+b_{n }x^{m+l}$

Evaluate $\lim_{x\to 0}\frac{p(x)}{q(x)}$ under the following conditions:

(i) $n=m$

(ii) $n>m,n-m$ is even

(iii) $n<m,n-m$ is even, $\frac{a_{0}}{b_{0}}>0$

(iv) $n<m,n-m$ is even, $\frac{a_{0}}{b_{0}}<0$

Why distinguish on whether $n-m$ is even and whether

$ \frac{a_0}{b_0} > 0\; \text{or}\; < 0?$

**pankaj** · June 16th 2009, 12:06 PM

Just think over it

Hint:what happens when you approach 0 from values less than 0.

**osodud** · June 16th 2009, 12:11 PM

Whats the final answer?. Im confused

**Danny** · June 16th 2009, 12:23 PM

Originally Posted by pankaj

Just think over it

Hint:what happens when you approach 0 from values less than 0.

Right, the possibility of approaching the vertical asymptotes $\pm \infty$ . My bad.

**pankaj** · June 16th 2009, 12:34 PM

Originally Posted by osodud

Whats the final answer?. Im confused

Give some time to it

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