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Math Help - problem's w/ l'hopital's rule

  1. #1
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    problem's w/ l'hopital's rule

    I have dificulty solving these problems:

    the first is:
    lim x -> 0 (from the left) ((3x-2)/(3x+2))^x

    I wasn't sure, could I just plug in 0 and get -1 instead of applying l'hopital's rule?

    the second is:
    lim x -> 1 (from the left) (x-1) (tan((pi * x)/2) )

    (tan((pi * x)/2) ) / (1/(x-1)
    since there is no multiplication rule. Took the derivative...

    ((sec(pi*x / 2)^2 * (pi/2)) / (-1/(x-1)^2)

    and this where i got really confused. Please help.
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  2. #2
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    I think the first one is indeterminate type \infty ^0. If lim f is of type 0^0, \infty^0 or 1^{\infty}, then lim(ln f) will be of the form 0 . \infty.
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  3. #3
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    Quote Originally Posted by Roam View Post
    I think the first one is indeterminate type \infty ^0. If lim f is of type 0^0, \infty^0 or 1^{\infty}, then lim(ln f) will be of the form 0 . \infty.
    The limit of the first one is actually of the form (-1)^{0^+} so l'hopitals rule does not apply. The function is discontinuous at x=0 and the limit doesn't exist.
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  4. #4
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    Thanks makaan, that furthered my believe in my answer.

    in #2,i tried the prob. again...
    i did the l'hopital's (LH) rule and ( rearranged because of a lack of multiplication rule) and got,
    (sec(pi * x/2)^2 * (pi / 2))/ ( -1 / (x - 1)^2) ... which is undef.

    i did it again, this i rearranged ((x-1)^2 (pi/2))/ (-1 / sec(pi*x /2) and the denominator turn into -cos^2 and got...

    (2(x-1)(pi/2) / (sin (pi*x / 2))^2 , i plugged in the 1 and got 0/1 = 0. is this right?
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