Results 1 to 8 of 8

Math Help - find limits using L'Hopital rule and FTC

  1. #1
    Senior Member
    Joined
    Nov 2010
    From
    Hong Kong
    Posts
    255

    find limits using L'Hopital rule and FTC

    Please kindly check if there are any mistakes.

    1. Find \lim_{x\rightarrow+\infty}e^{-e^x}\int_0^x e^{e^t} dt

    Solution.
    I use Fundamental Theorem of Calculus (FTC) \frac{d}{dx}\int_0^x f(t)dt=f(x)-f(0) to simplify the integral and then L'Hopital rule to find the limit:

    \lim_{x\rightarrow\infty}e^{-e^x}\int_0^x e^{e^t}dt=\lim_{x\rightarrow\infty}\frac{e^{e^x}-e^{e^0}}{e^{e^x}e^x}=\lim_{x\rightarrow\infty}\fra  c {e^{e^x}-e^{e^0}}{e^{e^x}e^x}=

    =\lim_{x\rightarrow\infty}\frac{e^{e^x}-e}{e^{e^x}e^x}=\lim_{x\rightarrow\infty}\frac{e^{e  ^x}(1-\frac{e}{e^{e^x}})}{e^{e^x}e^x}=\lim_{x\rightarrow  \infty}\frac{1-e/\infty}{e^x}=\lim_{x\rightarrow\infty}\frac{1-0}{\infty}=0



    2. Use L'Hopital's Rule to find the following limits:


    (i) lim_{x\rightarrow0}xlnx

    (ii) lim_{x\rightarrow0}x^x

    (iii) lim_{x\rightarrow0}\frac{1}{x}\int_0^x t^t dt


    Deduce the value of the limit lim_{x\rightarrow0}ln x \int_0^x t^t dt.


    Solution.

    (i) substitute y=\frac{1}{x} and take first derivative

    lim_{x\rightarrow0}xlnx=lim_{y\rightarrow\infty}\f  rac{ln1-lny}{y}=lim_{y\rightarrow\infty}\frac{(-ln y)}{y}=lim_{y\rightarrow\infty}(-\frac{1}{y})=0


    (ii) lim_{x\rightarrow0}x^x
    Take logarithm of the [tex]x^x{/math] and its limit:
    \ln(x^x)=xlnx and \lim_{x\rightarrow0} xlnx=0, see (i).
    Therefore if lim_{x\rightarrow0}\ln(x^x)=0, then the expression under log sign should be \rightarrow1.

    does this comment make sense?...



    (iii) \lim_{x\rightarrow0}\frac{1}{x}\int_0^x t^t dt

    Using L'Hopital Rule and FTC,


    =\lim_{x\rightarrow0}\frac{\frac{d}{dt}\int_0^xt^t  dt}{x}=\lim_{x\rightarrow0}(x^x-0^0)=1-1=0

    Here I am not sure at all whether I can take 0^0 - I rely on (ii) where I proved that lim_{x\rightarrow0}x^x=1.



    Therefore, lim_{x\rightarrow0}ln x \int_0^x t^t dt=\

    lim_{x\rightarrow0}lnx*\lim_{x\rightarrow0}\int_0^  x t^t dt=1*0=0

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by Volga View Post
    \frac{d}{dx}\int_0^x f(t)dt=f(x)-f(0)
    It should be:

    \dfrac{d}{dx}\displaystyle\int_0^x f(t)\;dt=f(x)



    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by Volga View Post
    lim_{x\rightarrow0}xlnx=lim_{y\rightarrow\infty}\f  rac{ln1-lny}{y}=lim_{y\rightarrow\infty}\frac{(-ln y)}{y}=lim_{y\rightarrow\infty}(-\frac{1}{y})=0
    Right.


    Therefore if lim_{x\rightarrow0}\ln(x^x)=0, then the expression under log sign should be \rightarrow1.
    does this comment make sense?...
    Better, suppose:

    L=\displaystyle\lim_{x \to a}{f(x)}\;,\quad\lambda=\displaystyle\lim_{x \to a}{\log f(x)}

    the, using the continuity of \log:

    e^{\lambda}=e^{\displaystyle\lim_{x \to a}{\log f(x)}}=e^{\log \left({\displaystyle\lim_{x \to a}{ f(x)}\right)}=\displaystyle\lim_{x \to a}{f(x)}=L


    =\lim_{x\rightarrow0}\frac{\frac{d}{dt}\int_0^xt^t  dt}{x}=\lim_{x\rightarrow0}(x^x-0^0)=1-1=0
    See my previous post. Besides, 0^0 is an indetermined expression.


    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2010
    From
    Hong Kong
    Posts
    255
    Quote Originally Posted by FernandoRevilla View Post
    It should be:

    \dfrac{d}{dx}\displaystyle\int_0^x f(t)\;dt=f(x)



    Fernando Revilla
    I am glad it is! This was my first version (I think I took it from one of the books) then I started having doubts and added f(0) by analogy with f(b)-f(a) when the limits of integration are a and b. I guess I need to work it out myself to stop having doubts.

    For 0^0, I googled for 0^0 and (1) google answered "1" and (2) there was this article which said "As a rule of thumb, one can say that 0^0 = 1".

    sci.math FAQ: What is 0^0?

    Conclusion: when it comes to mathematics, google is not a good source )))

    Thanks as always for your reply!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2010
    From
    Hong Kong
    Posts
    255
    Oh, is that because the derivative of a constant is zero?...

    So d/dt of F(0) is 0, no matter what function F is
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by Volga View Post
    For 0^0, I googled for 0^0 and (1) google answered "1" and (2) there was this article which said "As a rule of thumb, one can say that 0^0 = 1".
    When the expression 0^0 "comes from trying to find a limit" you can't say 0^0=1. You need additional information about the corresponding functions.


    Quote Originally Posted by Volga View Post
    Oh, is that because the derivative of a constant is zero?... So d/dt of F(0) is 0, no matter what function F is

    Please, look at the proof of the Fundamental Theorem of Calculus.


    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

  7. #7
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Just as a side note: there are way too many questions in the OP for one thread. See Rule # 8.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Nov 2010
    From
    Hong Kong
    Posts
    255
    Looked up the proof, no more questions. Thank you for your instruction!

    I will try not to keep to the limit of # of questions in a thread in the future. These ones all seemed to be related...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Limits using L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 11th 2011, 01:53 PM
  2. Replies: 1
    Last Post: April 24th 2010, 01:40 PM
  3. Using L'Hopital's Rule To Evaluate Limits
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 8th 2010, 02:37 AM
  4. Limits that L'Hopital's Rule doesnt work
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 2nd 2010, 02:47 PM
  5. limits(l'hopital's rule)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 16th 2008, 05:42 AM

Search Tags


/mathhelpforum @mathhelpforum