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Math Help - ln Limit

  1. #1
    Senior Member polymerase's Avatar
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    ln Limit

    How do you solve this limit, \displaystyle\lim_{x\to{0}}\frac{ln(\sqrt{x^5+3x+1  })}{x} without l'Hopital's Rule? Using l'Hopital's rule is too easy.
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  2. #2
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    Re: Limit

    First, you can use the properties of logarithms to take \ln (\sqrt{1+3x+x^5}) = \tfrac{1}{2}\ln (1+3x+x^5). Then, you can use the fact that for small x ( |x| \ll 1), \ln (1+x) \approx x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots (the Mercator series), and thus in the limit as x approaches zero, \ln (1+3x+x^5) approaches 3x-\tfrac{9}{2}x^2+\mathcal{O}(x^3), so thus
    \displaystyle\lim_{x\to{0}}\frac{ln(\sqrt{x^5+3x+1  })}{x} = \tfrac{1}{2}\displaystyle\lim_{x\to{0}}\frac{3x-\tfrac{9}{2}x^2+\cdots}{x} = \frac{3}{2}

    Hope this is illuminating.

    --Kevin C.
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by TwistedOne151 View Post
    First, you can use the properties of logarithms to take \ln (\sqrt{1+3x+x^5}) = \tfrac{1}{2}\ln (1+3x+x^5). Then, you can use the fact that for small x ( |x| \ll 1), \ln (1+x) \approx x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots (the Mercator series), and thus in the limit as x approaches zero, \ln (1+3x+x^5) approaches 3x-\tfrac{9}{2}x^2+\mathcal{O}(x^3), so thus
    \displaystyle\lim_{x\to{0}}\frac{ln(\sqrt{x^5+3x+1  })}{x} = \tfrac{1}{2}\displaystyle\lim_{x\to{0}}\frac{3x-\tfrac{9}{2}x^2+\cdots}{x} = \frac{3}{2}

    Hope this is illuminating.

    --Kevin C.
    Is there any other way...? I really unfamilar with the series (Mercator, taylor etc.) stuff.
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  4. #4
    GAMMA Mathematics
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    Quote Originally Posted by polymerase View Post
    Is there any other way...? I really unfamilar with the series (Mercator, taylor etc.) stuff.
    L'Hopital's rule is a nice trick to get past the tough work of using series and other definitions. BTW, why would you not want to use the rule?
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  5. #5
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    Re: Limit

    Not really. The presence of the logarithm ultimately requires either some form of series about the value to which its arguement converges, or else l'Hopital's rule (which essentially derives from the Taylor series for the numerator and denominator for the 0/0 indeterminate form). Ultimately you have to "know" in some fashion the behavior of ln(x) about x=1.
    Last edited by TwistedOne151; December 3rd 2007 at 09:36 AM. Reason: corrected for clarity
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  6. #6
    Senior Member polymerase's Avatar
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    Quote Originally Posted by colby2152 View Post
    L'Hopital's rule is a nice trick to get past the tough work of using series and other definitions. BTW, why would you not want to use the rule?
    If this was simply "find the answer" then of course i used that...but im trying to find "techniques" to imploy for hard limits because most of the time my prof won't let us use l'Hopital...when he does...the function is crazy.
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  7. #7
    Super Member PaulRS's Avatar
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    In many cases, like this one, it is enough to consider the following:

    \lim_{u\rightarrow{0}}\frac{\ln(u+1)}{u}=\lim_{u\r  ightarrow{0}}\ln(1+u)^{u}=1

    Because: \lim_{u\rightarrow{0}}(1+u)^{u}=e

    Thus: \lim_{x\rightarrow{0}}\frac{\ln(\sqrt[]{1+3x+5x^2})}{x}=\lim_{x\rightarrow{0}}\frac{\frac  {1}{2}\cdot{\ln(1+3x+5x^2)}}{x}=\frac{1}{2}\cdot{\  lim_{x\rightarrow{0}}\frac{\frac{\ln(1+3x+5x^2)}{3  x+5x^2}}{\frac{x}{3x+5x^2}}}

    \lim_{x\rightarrow{0}}\frac{\ln(\sqrt[]{1+3x+5x^2})}{x}=\frac{1}{2}\cdot{\lim_{x\rightarr  ow{0}}\frac{\frac{\ln(1+3x+5x^2)}{3x+5x^2}}{\frac{  x}{3x+5x^2}}}=\frac{1}{2}\cdot{\lim_{x\rightarrow{  0}}\frac{3x+5x^2}{x}}=\frac{3}{2}

    Since \frac{\ln(1+3x+5x^2)}{3x+5x^2}\rightarrow{1}
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  8. #8
    Senior Member polymerase's Avatar
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    Quote Originally Posted by PaulRS View Post
    In many cases, like this one, it is enough to consider the following:

    \lim_{u\rightarrow{0}}\frac{\ln(u+1)}{u}=\lim_{u\r  ightarrow{0}}\ln(1+u)^{u}=1

    Because: \lim_{u\rightarrow{0}}(1+u)^{u}=e

    Thus: \lim_{x\rightarrow{0}}\frac{\ln(\sqrt[]{1+3x+5x^2})}{x}=\lim_{x\rightarrow{0}}\frac{\frac  {1}{2}\cdot{\ln(1+3x+5x^2)}}{x}=\frac{1}{2}\cdot{\  lim_{x\rightarrow{0}}\frac{\frac{\ln(1+3x+5x^2)}{3  x+5x^2}}{\frac{x}{3x+5x^2}}}

    \lim_{x\rightarrow{0}}\frac{\ln(\sqrt[]{1+3x+5x^2})}{x}=\frac{1}{2}\cdot{\lim_{x\rightarr  ow{0}}\frac{\frac{\ln(1+3x+5x^2)}{3x+5x^2}}{\frac{  x}{3x+5x^2}}}=\frac{1}{2}\cdot{\lim_{x\rightarrow{  0}}\frac{3x+5x^2}{x}}=\frac{3}{2}

    Since \frac{\ln(1+3x+5x^2)}{3x+5x^2}\rightarrow{1}
    Nice
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    Nice
    Just be aware that, in general
    \lim_{x \to a}(f(u(x))) \neq f \left ( \lim_{x \to a}u(x) \right )

    It works for this problem, but it can't be generalized to all functions.

    -Dan
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  10. #10
    Super Member PaulRS's Avatar
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    Quote Originally Posted by topsquark View Post
    Just be aware that, in general
    \lim_{x \to a}(f(u(x))) \neq f \left ( \lim_{x \to a}u(x) \right )

    It works for this problem, but it can't be generalized to all functions.

    -Dan
    In this case it is true because of the continuity of the logarithm
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  11. #11
    Eater of Worlds
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    An interesting observation to make is that, in general,

    \lim_{x\rightarrow{0}}\frac{ln(ax^{2}+bx+1)}{cx}=\  frac{b}{c}

    In this case, b/c = 3/2.
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  12. #12
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    Quote Originally Posted by galactus View Post
    An interesting observation to make is that, in general,

    \lim_{x\rightarrow{0}}\frac{ln(ax^{2}+bx+1)}{cx}=\  frac{b}{c}

    In this case, b/c = 3/2.
    You should make some restrictions on the coefficients. Like a\not = 0 and that sort of stuff.
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