## anonymous 4 years ago lim(s->x) s^5-x^5/s^2-x^2 can someone please explain to me why this answer is not 0? i factored it out s^2-x^2 and canceled to get s^3-x^3

1. anonymous

because the only common factor of the numerator and denominator is $s-x$

2. anonymous

$s^5-x^5=(s-x) (s^4+s^3 x+s^2 x^2+s x^3+x^4)$

3. anonymous

ahhh wow im retard.... lol so how would you do it then?

4. anonymous

so you get that in the numerator, and $(s-x)(s+x)$ in the denominator, can only cancel the common factor

5. anonymous

got it

6. anonymous

you get $\frac{ (s^4+s^3 x+s^2 x^2+s x^3+x^4)}{s+x}$ after canceling, then replace x by s

7. anonymous

kk then factor out an x^2?

8. anonymous

x*

9. anonymous

nothing to factor now. your only job is $\lim_{s\rightarrow x}\frac{ (s^4+s^3 x+s^2 x^2+s x^3+x^4)}{s+x}$ you no longer have a zero in the denominator, so where you see an x, replace it by s you get a whole raft of $s^4$ in the numerator and $2s$ in the denominator. then you can cancel one of the "s"

10. anonymous

oh damn i had that backwards sorry replace the "s" by "x" etc.

11. anonymous

haha yeah i figured, thank you!@