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Math Help - Help with limits problems

  1. #1
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    Help with limits problems

    My problem says, "there's a myth circulating among calculus students which states that all indeterminate forms of types 0^0, infinity^0 and 1^(infinity) have a value of 1 since anything to the zero power is 1 and 1 to any power is 1.

    But the thing is, those 3 things I mentioned before are "descriptions of limits" rather than powers of numbers. Please show that such indeterminate forms can have any positive real value.

    (PS: Apologize for the terrible typing, I don't know how to use latex well.)

    1)

    lim (as x-> 0+)
    [x^(lna/(1+lnx))] = 0^0 = a

    2)

    lim (as x-> 0+)
    [x^(lna/(1+lnx))] = infinity^0 = a

    3)

    lim (as x-> 0+)
    [(x+1)^(lna/x)] = 1^(infinity) = a

    I am really stumped on how to do these. I did #1 and ended up with 1/(a(x+1)) which means that it would end up with a fraction of 1/a if x->0, which does not equal a.

    My attempt

    x^(lna/(1+lnx))

    I used the e^ln thing to get the exponent down. Then I used L'Hopital rule.

    e^(1/(a(x+1)))

    ?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    See attachment

    All are handled in basically the same way
    Attached Thumbnails Attached Thumbnails Help with limits problems-limit.jpg  
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  3. #3
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    Thanks! It seems like my problem was that I changed ln(a) to 1/a.

    Just wondering, now I know how to get y=a, how does the 0^0 thing relate to this problem?
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  4. #4
    MHF Contributor Calculus26's Avatar
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    lim (as x-> 0+)
    [x^(lna/(1+lnx))] = = a

    as x -> 0 ln(x) -> -inf

    ln(a)/-inf = 0

    so you have 0^0 as x->0
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  5. #5
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    Looks good, I understand now since x = 0 would be the base and then the 0 explained in the last post would be the exponent.

    How do I do this for part b and c where I need \infty^0 and 1^\infty?
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  6. #6
    MHF Contributor Calculus26's Avatar
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    2)

    lim (as x-> 0+)
    [x^(lna/(1+lnx))] = infinity^0 = a

    in #2 the limit should be as x-> infinty not zero

    It still evaluates to a

    in #3 its obvious x+1 ->1
    lna/x -> ln(a)/0 -> infinity
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  7. #7
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    That's weird, my problem definitely says that for all 3 parts the limit goes to 0. But I guess I'll explain that on the problem.

    Anyway, looks like another problem solved. Thanks.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Lord Darkin View Post
    My problem says, "there's a myth circulating among calculus students which states that all indeterminate forms of types 0^0, infinity^0 and 1^(infinity) have a value of 1 since anything to the zero power is 1 and 1 to any power is 1.

    But the thing is, those 3 things I mentioned before are "descriptions of limits" rather than powers of numbers. Please show that such indeterminate forms can have any positive real value.

    (PS: Apologize for the terrible typing, I don't know how to use latex well.)

    1)

    lim (as x-> 0+)
    [x^(lna/(1+lnx))] = 0^0 = a

    2)

    lim (as x-> 0+)
    [x^(lna/(1+lnx))] = infinity^0 = a

    3)

    lim (as x-> 0+)
    [(x+1)^(lna/x)] = 1^(infinity) = a

    I am really stumped on how to do these. I did #1 and ended up with 1/(a(x+1)) which means that it would end up with a fraction of 1/a if x->0, which does not equal a.

    My attempt

    x^(lna/(1+lnx))

    I used the e^ln thing to get the exponent down. Then I used L'Hopital rule.

    e^(1/(a(x+1)))

    ?
    Are these set problems, or can you show that 0^0 can converges to any value?
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  9. #9
    MHF Contributor Calculus26's Avatar
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    That's weird, my problem definitely says that for all 3 parts the limit goes to 0. But I guess I'll explain that on the problem.

    Anyway, looks like another problem solved. Thanks.


    Note if that were the case 1 and 2 are identical
    Last edited by Calculus26; January 18th 2010 at 06:54 PM.
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  10. #10
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    ^Yes, they are exactly identical. ??? My teacher won't take questions for xtra credit (e.g. I can't ask why they are the same) so I'll just hand in what I have.

    Thanks again to you guys.
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