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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

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  1. anonymous
    • 4 years ago
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    because the only common factor of the numerator and denominator is \[s-x\]

  2. anonymous
    • 4 years ago
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    \[s^5-x^5=(s-x) (s^4+s^3 x+s^2 x^2+s x^3+x^4)\]

  3. anonymous
    • 4 years ago
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    ahhh wow im retard.... lol so how would you do it then?

  4. anonymous
    • 4 years ago
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    so you get that in the numerator, and \[(s-x)(s+x)\] in the denominator, can only cancel the common factor

  5. anonymous
    • 4 years ago
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    got it

  6. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    kk then factor out an x^2?

  8. anonymous
    • 4 years ago
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    x*

  9. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    oh damn i had that backwards sorry replace the "s" by "x" etc.

  11. anonymous
    • 4 years ago
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    haha yeah i figured, thank you!@

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