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Math Help - L'Hopital's Rule and Improper Integrals

  1. #1
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    L'Hopital's Rule and Improper Integrals

    \lim (e^x+x)^\frac{2}{x}
    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.
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  2. #2
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    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
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  3. #3
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    Quote Originally Posted by saiyanmx89 View Post
    \lim (e^x+x)^\frac{2}{x}
    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

    \frac{0}{0} or \frac{\infty}{\infty}

    So which form do you get at the moment?
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  4. #4
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    The limit equals e^4. Here's my approach:

    \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 \left( 1+\frac{x}{e^{x}} \right)^{\frac{2}{x}}\to e^{2} as x\to0.

    We have \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 \ln \left( 1+\frac{x}{e^{x}} \right)\cdot \frac{e^{x}}{x}\to 1 as x\to0.

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

    Finally \left( 1+\frac{x}{e^{x}} \right)^{\frac{2}{x}}\to e^{2} as x\to0, as required and the whole limit tends to e^4.\quad\blacksquare
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  5. #5
    MHF Contributor matheagle's Avatar
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    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
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  6. #6
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    Quote Originally Posted by Prove It View Post
    Actually when dealing with limits, the indeterminate forms are

    \frac{0}{0} or \frac{\infty}{\infty}

    So which form do you get at the moment?
    0^0 ?
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  7. #7
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by saiyanmx89 View Post
    0^0 ?

    whener you have 0^0
    you just call that y=0^0
    and then proceed via \ln y...
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  8. #8
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    Quote Originally Posted by Prove It View Post
    Actually when dealing with limits, the indeterminate forms are

    \frac{0}{0} or \frac{\infty}{\infty}

    So which form do you get at the moment?
    0^0 and \infty^\infty are also "indeterminate forms" and, I think, more relevant here.
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