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Math Help - I Need Help Applying L'Hospital's Rule

  1. #1
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    I Need Help Applying L'Hospital's Rule

    I need help finding two limits. I already have the answers.

    Problem #1

    lim .... ln(x)
    x->0+ ----
    .......... x

    This appears to have a form of negative infinity over zero. I do not see how to simplify, or apply L'Hospital's rule.



    Problem #2

    lim ... x^a -a^x + a -1
    x->1 ----------------
    ............ (x-1)^2

    This has a form of 0/0, thus I apply L'Hospital's rule:

    = lim ... ax^(a-1) - a^x * ln(a)
    x->1 .. ------------------------
    ..................... 2(x-1)

    and I simplify:

    = lim .. ax^a - x* a^x * ln(a)
    x->1 . ------------------------
    .................... 2x(x-1)

    The denominator goes to zero, but the numerator goes to a - a * ln(a). Again I am stuck.

    Any help is appreciated.

    If someone could give me a concise algorithm which would help me always find the limits(at this level of math), I would be grateful.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by masonbrewey View Post
    I need help finding two limits. I already have the answers.

    Problem #1

    lim .... ln(x)
    x->0+ ----
    .......... x

    This appears to have a form of negative infinity over zero. I do not see how to simplify, or apply L'Hospital's rule.



    Problem #2

    lim ... x^a -a^x + a -1
    x->1 ----------------
    ............ (x-1)^2

    This has a form of 0/0, thus I apply L'Hospital's rule:

    = lim ... ax^(a-1) - a^x * ln(a)
    x->1 .. ------------------------
    ..................... 2(x-1)

    and I simplify:

    = lim .. ax^a - x* a^x * ln(a)
    x->1 . ------------------------
    .................... 2x(x-1)

    The denominator goes to zero, but the numerator goes to a - a * ln(a). Again I am stuck.

    Any help is appreciated.

    If someone could give me a concise algorithm which would help me always find the limits(at this level of math), I would be grateful.
    in the first one let
    y=ln(x) \Rightarrow e^y=x so when x\rightarrow 0^+ ....\;\;\quad y\rightarrow -\infty

    lim_{y\rightarrow -\infty}\frac{y}{e^y}  = lim_{y\rightarrow -\infty} y(e^{-y})

    the second one

    lim_{x\rightarrow 1}\frac{x^a-a^x+a-1}{(x-1)^2}=\frac{0}{0}=lop=lim_{x\rightarrow 1}\frac{ax^{a-1}-a^x(lna)}{2(x-1)}=\frac{a-aln(a)}{0}=\infty
    Last edited by Amer; June 29th 2009 at 04:37 AM.
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  3. #3
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    Amer,

    Thank you for your help. However, according to the answer sheet, the answer to the first is negative infinity and 1/2*a(a - 1) for the second.
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  4. #4
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    Quote Originally Posted by Amer View Post
    in the first one let
    y=ln(x) \Rightarrow e^y=x so when x\rightarrow 0^+ ....\;\;\quad y\rightarrow +\infty

    lim_{y\rightarrow +\infty}\frac{y}{e^y} = \frac{+\infty}{+\infty} =lop=lim_{y\rightarrow +\infty}\frac{1}{e^y}=0

    the second one

    lim_{x\rightarrow 1}\frac{x^a-a^x+a-1}{(x-1)^2}=\frac{0}{0}=lop=lim_{x\rightarrow 1}\frac{ax^{a-1}-a^x(lna)}{2(x-1)}=\frac{a-aln(a)}{0}=\infty
    I believe you'll find that \lim_{x \to 0^{+}}{\ln{x}} = -\infty...
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  5. #5
    MHF Contributor Amer's Avatar
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    sorry ...
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