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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because the only common factor of the numerator and denominator is \[sx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[s^5x^5=(sx) (s^4+s^3 x+s^2 x^2+s x^3+x^4)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahhh wow im retard.... lol so how would you do it then?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so you get that in the numerator, and \[(sx)(s+x)\] in the denominator, can only cancel the common factor

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0kk then factor out an x^2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nothing 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"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh damn i had that backwards sorry replace the "s" by "x" etc.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha yeah i figured, thank you!@
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