Results 1 to 8 of 8

Thread: L'Hopital's Rule and Improper Integrals

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    100

    L'Hopital's Rule and Improper Integrals

    $\displaystyle \lim (e^x+x)^\frac{2}{x}$
    $\displaystyle x->0^+$

    Describe the type of indeterminate form that is obtained by direct substitutions.
    Evaluate the integral using L'Hopital's Rule if necessary.

    I understand that indeterminate means having no definite or definable value. But I don't what to put for it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2009
    Posts
    36
    lt(x-->0+)(e^x+x)^2/x

    p=(e^x+x)^2/x

    so,ln p=2ln(e^x+x)/x

    so,lim ln p=2 lim (x-->0+)ln (e^x+x)/x

    so,ln lim p=2 lim(x-->0+)1/(e^x+x)[by l'hospital's rule]

    so,lim p=e^2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by saiyanmx89 View Post
    $\displaystyle \lim (e^x+x)^\frac{2}{x}$
    $\displaystyle x->0^+$

    Describe the type of indeterminate form that is obtained by direct substitutions.
    Evaluate the integral using L'Hopital's Rule if necessary.

    I understand that indeterminate means having no definite or definable value. But I don't what to put for it.
    Actually when dealing with limits, the indeterminate forms are

    $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$

    So which form do you get at the moment?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    The limit equals $\displaystyle e^4.$ Here's my approach:

    $\displaystyle \left( e^{x}+x \right)^{\frac{2}{x}}=\left\{ e^{x}\bigg( 1+\frac{x}{e^{x}} \bigg) \right\}^{\frac{2}{x}}=e^{2}\left( 1+\frac{x}{e^{x}} \right)^{\frac{2}{x}},$ thus we only need to prove that $\displaystyle \left( 1+\frac{x}{e^{x}} \right)^{\frac{2}{x}}\to e^{2}$ as $\displaystyle x\to0.$

    We have $\displaystyle \left( 1+\frac{x}{e^{x}} \right)^{\frac{2}{x}}=\exp \left\{ \frac{2}{e^{x}}\cdot \ln \bigg( 1+\frac{x}{e^{x}} \bigg)\cdot \frac{e^{x}}{x} \right\},$ and $\displaystyle \ln \left( 1+\frac{x}{e^{x}} \right)\cdot \frac{e^{x}}{x}\to 1$ as $\displaystyle x\to0.$

    (Just we need to recall that $\displaystyle \underset{x\to 0}{\mathop{\lim }}\,\frac{\ln (1+x)}{x}=1.$)

    Finally $\displaystyle \left( 1+\frac{x}{e^{x}} \right)^{\frac{2}{x}}\to e^{2}$ as $\displaystyle x\to0,$ as required and the whole limit tends to $\displaystyle e^4.\quad\blacksquare$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    e^4 is the answer
    the error here is with the chain rule...
    you were missing the e^x+1--->2 as x-->0
    which is why you only got e^2 and not e^4.


    Quote Originally Posted by sbcd90 View Post
    lt(x-->0+)(e^x+x)^2/x

    p=(e^x+x)^2/x

    so,ln p=2ln(e^x+x)/x

    so,lim ln p=2 lim (x-->0+)ln (e^x+x)/x

    so,ln lim p=2 lim(x-->0+)(e^x+1)/(e^x+x)[by l'hospital's rule]

    so,lim p=e^[(2)(2)]=e^4
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jan 2009
    Posts
    100
    Quote Originally Posted by Prove It View Post
    Actually when dealing with limits, the indeterminate forms are

    $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$

    So which form do you get at the moment?
    $\displaystyle 0^0$ ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Quote Originally Posted by saiyanmx89 View Post
    $\displaystyle 0^0$ ?

    whener you have $\displaystyle 0^0$
    you just call that $\displaystyle y=0^0$
    and then proceed via $\displaystyle \ln y$...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,775
    Thanks
    3028
    Quote Originally Posted by Prove It View Post
    Actually when dealing with limits, the indeterminate forms are

    $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$

    So which form do you get at the moment?
    $\displaystyle 0^0$ and $\displaystyle \infty^\infty$ are also "indeterminate forms" and, I think, more relevant here.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. l'HŰpital's rule
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 24th 2011, 06:25 AM
  2. Improper Integral's & L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Feb 22nd 2011, 04:25 AM
  3. L'hopital's rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 30th 2008, 08:35 AM
  4. l'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 21st 2007, 08:59 AM
  5. líHopitalís rule
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Dec 2nd 2006, 05:21 AM

Search Tags


/mathhelpforum @mathhelpforum