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Math Help - Find the limit

  1. #1
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    Find the limit

    Question : Find the Limite \lim_{x \to 0} (e^x + x)^{\frac{1}{x}}


    My Work :::::::::::::::::::::::::::::::::::::::::::

    ((e^x - 1) + (1+x))^{\frac{1}{x}}.....................Is this step correct
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  2. #2
    MHF Contributor matheagle's Avatar
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    As, you usually say to me, I have no idea what you are doing.
    The idea here is to take the logarithm and then use L'Hopital's rule.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zorro View Post
    Question : Find the Limite \lim_{x \to 0} (e^x + x)^{\frac{1}{x}}


    My Work :::::::::::::::::::::::::::::::::::::::::::

    ((e^x - 1) + (1+x))^{\frac{1}{x}}.....................Is this step correct
    Two main methods:

    Method one: \lim_{x\to\infty}\left(e^x+x\right)^{\frac{1}{x}}=  \lim_{x\to\infty}e\left(1+\frac{x}{e^x}\right)^{\f  rac{1}{x}} and it is quite easy to validate that \lim_{x\to\infty}\left(1+\frac{x}{e^x}\right)^{\fr  ac{1}{x}}=\exp\left(\lim_{x\to\infty}\frac{\ln\lef  t(1+\tfrac{x}{e^x}\right)}{x}\right) and since \ln\left(1+\tfrac{x}{e^x}\right)\underset{x\to\inf  ty}{\sim}\tfrac{x}{e^x} it follows that our limit is equal to \exp\left(\lim_{x\to\infty}\frac{\tfrac{x}{e^x}}{x  }\right)=e^0=1 so then the actual limit is clearly e.


    Alternatively,

    e\leqslant\left(e^x+x\right)^{\frac{1}{x}}\leqslan  t 2^{\frac{1}{x}}e. And since \lim_{x\to\infty}2^{\frac{1}{x}}=1 the limit follows by the squeeze theorem.
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  4. #4
    MHF Contributor matheagle's Avatar
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    I just let

    y=\lim_{x \to 0} (e^x + x)^{\frac{1}{x}}

    then take the log of y, pass through the limit and apply L'Hopital rule's.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by matheagle View Post
    I just let

    y=\lim_{x \to 0} (e^x + x)^{\frac{1}{x}}

    then take the log of y, pass through the limit and apply L'Hopital rule's.
    That is of course equally valid!
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  6. #6
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    y = (e^x + x)^\frac{1}{x}

    taking log we get

    = \frac{ln (e^x+x)}{x}

    Using l'Hospital rule

    f'(x) = \frac{d}{dx} \left[ ln(e^x + x) \right] = \frac{e^x + 1}{e^x+x}

    g'(x) = 1

    \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \ \frac{e^x+1}{e^x+x} .............what should i do know???
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  7. #7
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    Quote Originally Posted by Drexel28 View Post



    <br />
\lim_{x\to\infty}\left(e^x+x\right)^{\frac{1}{x}}=  \lim_{x\to\infty}e\left(1+\frac{x}{e^x}\right)^{\f  rac{1}{x}}<br />
    A nice method !! How could you think of this ...

    You did the limit for  x \to \infty ,
    now if  x \to 0 , let me finish it

     \lim_{x\to 0} e \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}

     = e \lim_{x\to 0} \left( 1 + \frac{ e^{-x}}{ \frac{1}{x}} \right)^{\frac{1}{x}} = e \lim_{x\to 0} e^{e^{-x}} = e^2
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  8. #8
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    Here is another method to reach the limit but  x \to \infty


    starting from

     e \lim_{x\to \infty} \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}

    which is the result from Drexel28 .


    since  \ln(u) = \lim_{a\to 0} \frac{ u^a - 1}{a}

    \lim_{x\to \infty} \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}

     = \lim_{x\to\infty} \frac{1}{x} \cdot \frac{\left [ \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}} - 1 \right ]}{ \frac{1}{x}}  + 1


    sub  \frac{1}{x} = a

     = \lim_{a\to 0} a \ln( 1 + \frac{{e^{- \frac{1}{a}}}}{a} ) + 1

    we can see that it equals to  1

    (  0 * \ln(1) + 1 )
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by simplependulum View Post
    A nice method !! How could you think of this ...

    You did the limit for  x \to \infty ,
    now if  x \to 0 , let me finish it

     \lim_{x\to 0} e \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}

     = e \lim_{x\to 0} \left( 1 + \frac{ e^{-x}}{ \frac{1}{x}} \right)^{\frac{1}{x}} = e \lim_{x\to 0} e^{e^{-x}} = e^2
    Another really easy thing one can prove is that let L=\left(P(x)+\alpha^x+\beta^x+\cdots\zeta^x\right)  ^{\frac{1}{x}} be such that P(x) is an arbitrary polynomial and 0<\alpha<\beta<\cdots<\zeta then \ln\text{ }L=\lim_{x\to\infty}\frac{\ln\left(P(x)+\alpha^x+\  beta^x+\cdots+\zeta^x\right)}{x} =\lim_{x\to\infty}\left\{\ln\left(\zeta\right)+\fr  ac{\ln\left(\tfrac{P(x)}{\zeta^x}+\left(\tfrac{\al  pha}{\zeta}\right)^x+\left(\tfrac{\beta}{\zeta}\ri  ght)+<br />
\cdots+1\right)}{x}\right\} and since \frac{P(x)}{\zeta^x}+\left(\frac{\alpha}{\zeta}\ri  ght)^x+\left(\frac{\beta}{\zeta}\right)^x+\cdots\t  o0 we can see that \ln\left(\frac{P(x)}{\zeta^x}+\left(\tfrac{\alpha}  {\zeta}\right)^x+\left(\tfrac{\beta}{\zeta}\right)  ^x+\cdots+1\right)\underset{x\to\infty}{\sim}\frac  {P(x)}{\zeta^x}+\left(\frac{\alpha}{\zeta}\right)^  x+\left(\frac{\beta}{\zeta}\right)^x+\cdots from where it clearly follows that \ln\text{ }L=\ln\left(\zeta\right)\implies L=\zeta. A similar proof can be done wit h the squeeze theorem.


    This can be GREATLY generalized.
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  10. #10
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    shouldnt it be this

    Quote Originally Posted by simplependulum View Post
    A nice method !! How could you think of this ...

    You did the limit for  x \to \infty ,
    now if  x \to 0 , let me finish it

     \lim_{x\to 0} e \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}


    I am no genuis but should it be this

     \lim_{x\to 0} e^x \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}

    u have taken e out side instead of e^x
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  11. #11
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    Quote Originally Posted by zorro View Post
    I am no genuis but should it be this

     \lim_{x\to 0} e^x \left( 1 + \frac{x}{e^x} \right)^{\frac{1}{x}}

    u have taken e out side instead of e^x
    Yes, you are right. It was probably a typo.
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  12. #12
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    Quote Originally Posted by zorro View Post
    y = (e^x + x)^\frac{1}{x}

    taking log we get

    = \frac{ln (e^x+x)}{x}

    Using l'Hospital rule

    f'(x) = \frac{d}{dx} \left[ ln(e^x + x) \right] = \frac{e^x + 1}{e^x+x}

    g'(x) = 1

    \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \ \frac{e^x+1}{e^x+x} .............what should i do know???
    Now just set x= 0!
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