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calculus

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim x→0 (e^x − e^−x − 2x)/(x − sin(x))

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christina

  1. you do have a 0/0 form

    Using Hospital's rule
    then lim= (e^x+e^-x-2)/(x+cosx)

    Lim= 0/1=0

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    bobpursley

  2. using L'Hospital's rule we have

    lim x→0 (e^x − e^−x − 2x)/(x − sin(x))
    lim x→0 (e^x + e^−x − 2)/(1 − cosx)
    still 0/0, so do it again:
    lim x→0 (e^x - e^−x)/sinx
    and again:
    lim x→0 (e^x + e^−x)/cosx
    → (1+1)/1 = 2

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    Steve

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