Results 1 to 8 of 8

Math Help - another limit

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    81

    another limit

    \displaystyle \lim_{x \to 0^{+}} x^{1/3} \ln x =

    How do I go about solving this kind of limit?
    Do I handle each part individually?

    The only thing I can think about, is taking the derivative of each part.

    <br />
\displaystyle \frac{1}{3x^{\frac{2}{3}}} * \frac{1}{x}<br />

    To me, this still looks like it is undefined..
    Can someone please explain how to do these types? Perhaps a link to a tutorial too. I'm not sure what to search online for these types either. Do they have a special name?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle\lim f*g=\lim f *\lim g

    \displaystyle \lim_{x\to 0}x^{\frac{1}{3}}=0

    \displaystyle 0*\lim_{x\to 0}ln(x)=0

    A useful tool for some limits is:

    \displaystyle \lim_{x\to\infty}f(x)=L \ \mbox{iff.} \lim_{x\to 0}\frac{1}{f(x)}=L

    Using this on your limit, we get:

    \displaystyle \lim_{x\to\infty}\frac{1}{x^{\frac{1}{3}}*ln(x)}=\  lim_{x\to\infty}\frac{1}{\infty}=0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2010
    Posts
    81
    Thank you!

    So, since the limit of f(x) =0, we can ignore whatever the limit of g(x) is, since we are multiplying. But for this scenario, would it be safe to say that the limit of g(x) would have been undefined? In that case 0 * undefined = 0?

    Thanks for that tool!
    However, did you mean to say:
    <br />
\displaystyle \lim_{x\to 0}f(x)=L \ \mbox{iff.} \lim_{x\to \infty}\frac{1}{f(x)}=L<br />

    did you have the 0 & infinity symbols backwards?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    You can't just ignore anything and no the 0 and infinity aren't backwards.

    For example, lets consider the function \displaystyle f(x)=\frac{1}{x} then the \displaystyle\lim_{x\to\infty}\frac{1}{x}=0

    Ok so lets use the definition now. \displaystyle\frac{1}{f(x)}=\frac{1}{\frac{1}{x}}=  x

    \displaystyle\lim_{x\to 0}{x}=0
    Last edited by dwsmith; November 11th 2010 at 02:48 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2010
    Posts
    81
    Ok IC. So for
    <br />
\displaystyle \lim_{x \to 1^{+}} ( \displaystyle \frac {1}{\ln x} - \displaystyle \frac {1}{x - 1} ) =<br />

    It would be the same as: Lim f(x) - Lim f(g)
    so for:
    <br />
\lim_{x \to 1^{+}}\frac{1}{lnx} = \infty<br />
    x becomes really close to one, making ln(x) extremely small. Then 1/rly small = rly big and approaches infinity?

    Same with the other term:
    [Math]
    \displaystyle \lim_{x \to 1^{+}} \frac{1}{x-1} = \infty
    [/tex]

    because x-1 will approach a super small number, and the reciprocal of that will be a rly big number, approaching infinity.

    So, then we have infinity-infinity = 0?

    But this appears to be wrong. What is wrong with my reasoning?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    You can't subtract infinities. Some infinities are bigger than others. You need to use L'Hopital's Rule here.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,683
    Thanks
    446
    Quote Originally Posted by Vamz View Post
    Ok IC. So for
    <br />
\displaystyle \lim_{x \to 1^{+}} ( \displaystyle \frac {1}{\ln x} - \displaystyle \frac {1}{x - 1} ) =<br />

    It would be the same as: Lim f(x) - Lim f(g)
    so for:
    <br />
\lim_{x \to 1^{+}}\frac{1}{lnx} = \infty<br />
    x becomes really close to one, making ln(x) extremely small. Then 1/rly small = rly big and approaches infinity?

    Same with the other term:
    [Math]
    \displaystyle \lim_{x \to 1^{+}} \frac{1}{x-1} = \infty
    [/tex]

    because x-1 will approach a super small number, and the reciprocal of that will be a rly big number, approaching infinity.

    So, then we have infinity-infinity = 0?

    But this appears to be wrong. What is wrong with my reasoning?
    \infty - \infty is an indeterminate form.

    \displaystyle \frac{1}{\ln{x}} - \frac{1}{x-1}

    \displaystyle \frac{(x-1) - \ln{x}}{\ln{x}(x-1)}

    use L'Hopital ...

    \displaystyle \frac{1 - \frac{1}{x}}{\ln{x} + \frac{x-1}{x}}

    multiply numerator and denominator by x ...

    \displaystyle \frac{x - 1}{x\ln{x} + (x-1)}

    use L'Hopital again ...

    \displaystyle \frac{1}{2 + \ln{x}}

    limit as x \to 1^+ is \frac{1}{2}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Apr 2010
    Posts
    81
    There we go. I would have never thought of multiplying x on both the numerator and denominator :s


    thanks a ton!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 8th 2010, 11:29 AM
  2. Replies: 1
    Last Post: February 5th 2010, 03:33 AM
  3. Replies: 16
    Last Post: November 15th 2009, 04:18 PM
  4. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 05:05 PM
  5. Replies: 15
    Last Post: November 4th 2007, 07:21 PM

Search Tags


/mathhelpforum @mathhelpforum