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Thread: finding a limit

  1. October 4th 2009, 05:42 AM #1
    collegestudent321
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    finding a limit

    hi guys...
    Im having difficulty finding this limit and i wanted to know if anyone can help me out:

    lim e^(x)/ (1+(e^(2x))^(3/2))
    x -> infiniti


    I tried using L'Hospital's rule but that only made it more difficult to see. I feel like there's an easy trick that im just forgetting. If anyone could help me out that would be great. Thanks!
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  2. October 4th 2009, 05:51 AM #2
    DeMath
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    Quote Originally Posted by collegestudent321 View Post
    hi guys...
    Im having difficulty finding this limit and i wanted to know if anyone can help me out:

    lim e^(x)/ (1+(e^(2x))^(3/2))
    x -> infiniti


    I tried using L'Hospital's rule but that only made it more difficult to see. I feel like there's an easy trick that im just forgetting. If anyone could help me out that would be great. Thanks!
    Not difficult

    \mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}<br /> {{{{\left( {1 + {e^{2x}}} \right)}^{3/2}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}<br /> {{{e^{3x}}{{\left( {\frac{1}<br /> {{{e^{2x}}}} + 1} \right)}^{3/2}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}<br /> {{{e^{2x}}{{\left( {\frac{1}<br /> {{{e^{2x}}}} + 1} \right)}^{3/2}}}} = 0.
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  3. October 4th 2009, 06:30 AM #3
    collegestudent321
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    oh wow, ok i see. haha, that was easy. Thank you!!!
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  4. October 4th 2009, 06:32 AM #4
    collegestudent321
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    no wait... why is it e^(3x)????
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  5. October 4th 2009, 06:46 AM #5
    DeMath
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    Quote Originally Posted by collegestudent321 View Post
    no wait... why is it e^(3x)????
    Because

    \begin{gathered}<br /> {\left( {1 + {e^{2x}}} \right)^{3/2}} = {\left( {\sqrt {1 + {e^{2x}}} } \right)^3} = {\left( {{e^x}\sqrt {\frac{1}<br /> {{{e^{2x}}}} + 1} } \right)^3} = \hfill \\<br /> = {e^{3x}}{\left( {\sqrt {\frac{1}<br /> {{{e^{2x}}}} + 1} } \right)^3} = {e^{3x}}{\left( {\frac{1}<br /> {{{e^{2x}}}} + 1} \right)^{3/2}}. \hfill \\ <br /> \end{gathered}
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