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Math Help - Limit (no hospital)

  1. #1
    Super Member dhiab's Avatar
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    Limit (no hospital)

    Calculate :
     <br />
\lim_{x\rightarrow +\infty }(e^{x}-1)^{\frac{1}{x}}<br />
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  2. #2
    MHF Contributor ebaines's Avatar
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    Substitute y = 1/x, and then evaluate:

    <br />
\lim_ {y \to 0} (e ^ {1/y} -1) ^ y<br />

    You can then expand:

    <br />
(e ^ {1/y} -1) ^ y = (e^ {1/y} )^y - y (e^ {1/y} )^{y-1} + \frac { y (y-1)} 2 (e^ {1/y} )^{y-2} - ... <br />
     = e^1 - y e ^ {(\frac {y-1} y )} + \frac { y (y-1)} 2 (e^ {\frac {y-2} y )} - ...<br />

    You can see that as  y \to 0, you're left with just e.
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  3. #3
    MHF Contributor red_dog's Avatar
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    \lim_{x\to\infty}(e^x-1)^{\frac{1}{x}}=\lim_{x\to\infty}e^{\displaystyle  \ln(e^x-1)^{\frac{1}{x}}}=e^{\displaystyle\lim_{x\to\infty  }\frac{\ln(e^x-1)}{x}}

    Now, \frac{\ln(e^x-1)}{x}=\frac{\ln\left[e^x\left(1-\frac{1}{e^x}\right)\right]}{x}=\frac{x+\ln\left(1-\frac{1}{e^x}\right)}{x}=

    =1+\frac{\ln\left(1-\frac{1}{e^x}\right)}{-\frac{1}{e^x}}\cdot\left(-\frac{1}{xe^x}\right)\to 1+1\cdot 0=1

    So the limit is e.
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  4. #4
    Super Member dhiab's Avatar
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    I'have anthor solution :
     <br />
\lim_{X\to+ \infty }(e^{x}-1)^{\frac{1}{x}}=\lim_{x\to +\infty }(e^{x}(1-e^{-x}))^{\frac{1}{x}}<br />
     <br />
\lim_{x\to +\infty }(e^{x})^{\frac{1}{x}}(1-e^{-x})^{\frac{1}{x}}=\lim_{x\to +\infty }e(1-e^{-x})^{\frac{1}{x}}=e\times 1=e<br />
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