1. ## Polinomials and Limit

Hello

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

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

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

and

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

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

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Now consider
$\displaystyle p(x)=a_{0}x^n+a_{1}x^{n+1}+a_{2}x^{n+2}2+.....+a_{ n}x^{n+k}$

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

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

(i)$\displaystyle n=m$

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

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

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

3. Originally Posted by pankaj
$\displaystyle p(x)=a_{0}+a_{1}x+a_{2}x^2+.....+a_{n}x^n$

and

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

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

.................................................. ...............................................
Now consider
$\displaystyle p(x)=a_{0}x^n+a_{1}x^{n+1}+a_{2}x^{n+2}2+.....+a_{ n}x^{n+k}$

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

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

(i)$\displaystyle n=m$

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

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

(iv)$\displaystyle n<m,n-m$ is even,$\displaystyle \frac{a_{0}}{b_{0}}<0$
Why distinguish on whether $\displaystyle n-m$ is even and whether

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

4. Just think over it
Hint:what happens when you approach 0 from values less than 0.

5. Whats the final answer?. Im confused

6. 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 $\displaystyle \pm \infty$. My bad.

7. Originally Posted by osodud
Whats the final answer?. Im confused
Give some time to it