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Math Help - limits(l'hopital's rule)

  1. #1
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    limits(l'hopital's rule)

    lim_{x\rightarrow{\infty}}(\frac{x+1}{x})^x therefore i applied l'hopital s rule to make  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?
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  2. #2
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    Quote Originally Posted by ashes View Post
    lim_{x\rightarrow{\infty}}(\frac{x+1}{x})^x therefore i applied l'hopital s rule to make  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:


    \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.
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  3. #3
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    Quote Originally Posted by ashes View Post
    lim_{x\rightarrow{\infty}}(\frac{x+1}{x})^x therefore i applied l'hopital s rule to make  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 x^x via basic powers rule and get x*x^{x-1}. Notice that the result simplifies back to x^x

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

    Start with:

    y=x^x

    Take natural log of both sides...

    \ln|y|=x \ln|x|

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

    Derive with respect to x on both sides

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

    Substitute for y from original equation

    y' = x^x(\ln|x| + 1)
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  4. #4
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    Interestingly we were taught to do this with a different perspective.
    d(y^x) = x.y^{(x-1)} dy + y^x \ln |y| dx

    Now substitute y=x

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

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

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