# Thread:

1. ## 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!

2. Originally Posted by collegestudent321
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

$\mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}<br /> {{{{\left( {1 + {e^{2x}}} \right)}^{3/2}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}<br /> {{{e^{3x}}{{\left( {\frac{1}<br /> {{{e^{2x}}}} + 1} \right)}^{3/2}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}<br /> {{{e^{2x}}{{\left( {\frac{1}<br /> {{{e^{2x}}}} + 1} \right)}^{3/2}}}} = 0.$

3. oh wow, ok i see. haha, that was easy. Thank you!!!

4. no wait... why is it e^(3x)????

5. Originally Posted by collegestudent321
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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