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(1 - 5/x)^x with lim ->infinity
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is it e^5?
as x gets large; 5/x goes to zero; which leaves 1^inf .... is say it goes to 1
well, wolfram says it e^(-5)
thank you!
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\(\infty^0\) is one of the standard indeterminate forms. For this you must employ l'Hopital's rule. http://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms
\[\lim_{x\to \infty}\left(1+\frac{r}{x}\right)^{xt}=e^{rt}\] replace r by -5 and t by 1
Yeah, if you know all the different forms of e you can just do it by sight ;)
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