Results 1 to 4 of 4

Thread: limits(l'hopital's rule)

  1. #1
    Junior Member
    Joined
    Mar 2008
    Posts
    60

    limits(l'hopital's rule)

    $\displaystyle lim_{x\rightarrow{\infty}}(\frac{x+1}{x})^x$ therefore i applied l'hopital s rule to make $\displaystyle lim_{x\rightarrow{\infty}}\frac{-(x+1)^{x-1}}{x^2}$ I hope my differentiation is correct. So do i keep differentiating till I remove the x term from the denominator before i apply the limit?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by ashes View Post
    $\displaystyle lim_{x\rightarrow{\infty}}(\frac{x+1}{x})^x$ therefore i applied l'hopital s rule to make $\displaystyle lim_{x\rightarrow{\infty}}\frac{-(x+1)^{x-1}}{x^2}$ I hope my differentiation is correct. So do i keep differentiating till I remove the x term from the denominator before i apply the limit?
    I'm sorry but you're differentiations are terribly wrong. If you show how you did them I'll point out the many salto mortales .....

    You actually have a standard form:


    $\displaystyle \lim_{x\rightarrow{\infty}}\left(\frac{x+1}{x}\rig ht)^x = \lim_{x\rightarrow{\infty}}\left(1 + \frac{1}{x}\right)^x = e$


    by definition.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Thanks
    1
    Awards
    1
    Quote Originally Posted by ashes View Post
    $\displaystyle lim_{x\rightarrow{\infty}}(\frac{x+1}{x})^x$ therefore i applied l'hopital s rule to make $\displaystyle lim_{x\rightarrow{\infty}}\frac{-(x+1)^{x-1}}{x^2}$ I hope my differentiation is correct. So do i keep differentiating till I remove the x term from the denominator before i apply the limit?
    Mr. Fantastic's answer is correct, but I have one point to make...

    All too common mistake... you cannot derive $\displaystyle x^x$ via basic powers rule and get $\displaystyle x*x^{x-1}$. Notice that the result simplifies back to $\displaystyle x^x$

    You need to use the natural log and implicit differentiation to solve for such a derivative.

    Start with:

    $\displaystyle y=x^x$

    Take natural log of both sides...

    $\displaystyle \ln|y|=x \ln|x|$

    $\displaystyle \frac{1}{y}y' = \ln|x| + 1$

    Derive with respect to x on both sides

    $\displaystyle y' = y(\ln|x| + 1)$

    Substitute for y from original equation

    $\displaystyle y' = x^x(\ln|x| + 1)$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Interestingly we were taught to do this with a different perspective.
    $\displaystyle d(y^x) = x.y^{(x-1)} dy + y^x \ln |y| dx$

    Now substitute y=x

    $\displaystyle d(x^x) = x.x^{(x-1)} dx + x^x \ln |x| dx$

    $\displaystyle d(x^x) = x^{x} dx + x^x \ln |x| dx$

    $\displaystyle \frac{d(x^x)}{dx} = x^{x} + x^x \ln |x|$
    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: Dec 11th 2011, 01:53 PM
  2. find limits using L'Hopital rule and FTC
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Jan 13th 2011, 05:05 PM
  3. Replies: 1
    Last Post: Apr 24th 2010, 01:40 PM
  4. Using L'Hopital's Rule To Evaluate Limits
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Apr 8th 2010, 02:37 AM
  5. Limits that L'Hopital's Rule doesnt work
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 2nd 2010, 02:47 PM

Search Tags


/mathhelpforum @mathhelpforum