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RONNCC
 one year ago
Best ResponseYou've already chosen the best response.0ok so what do you think? :)

RONNCC
 one year ago
Best ResponseYou've already chosen the best response.0the goal is to help, not to tell you the answer. sorry, we can't help on your test ;). So look up l'hopital and indeterminate form next.

RONNCC
 one year ago
Best ResponseYou've already chosen the best response.0yup. look up l'hopital and "indeterminate" form.

Azteck
 one year ago
Best ResponseYou've already chosen the best response.1Dude Hero gave it away man. 1/x vanishes and then you have \[(1)^{x}\]

RONNCC
 one year ago
Best ResponseYou've already chosen the best response.0@Azteck how did he give it away? . ok first get @Argos to find out what indeterminate form is ____

Argos
 one year ago
Best ResponseYou've already chosen the best response.0i know inderminite form is 0/0 or inf/inf

Argos
 one year ago
Best ResponseYou've already chosen the best response.0and can to use lhospitals rule on those but nega exponent

Azteck
 one year ago
Best ResponseYou've already chosen the best response.1Then you get this: \[\frac{ 1 }{ (1){^x} }\] \[\frac{ 1 }{ (1)^{\infty} }\] \[=\frac{ 1 }{ 1 }\]

Argos
 one year ago
Best ResponseYou've already chosen the best response.0because the back of the book sais its e

Argos
 one year ago
Best ResponseYou've already chosen the best response.0nope I even doubel checked on wolfram http://www.wolframalpha.com/input/?i=lim+as+x%3E+inf+of+%281%281%2Fx%29%29^x

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.3Limit x approaches to infinity ( 1+ (1/x) )^x = e In your case, just replace x by x in this formula. So, it's e.

binarymimic
 one year ago
Best ResponseYou've already chosen the best response.0yeah 1 to the infinity is indeterminate so substitution will not work

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0shouldnt answer be 2 ? from binomial approximation ?

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.3Shouldn't it be e? A more accurate answer?

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0i can see easily why e fits here.. but i dont understand why doesnt the ans come from binomial approximation

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.3Sorry, I don't know binomial...

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0ahh,maybe i see why.. power should be a real no. for that, here, power is not a real no.

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0dw:1357649297414:dw

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.3Even though we use L'Hopital's Rule, we get the same answer.. Let \(y= (1\frac{1}{x})^{x}\) \[\ln y= x \ln (1\frac{1}{x})\]\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}x \ln (1\frac{1}{x})\]\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{ \ln (1\frac{1}{x})}{\frac{1}{x}}\]Use L'Hopital's Rule:\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{\frac{1}{{(1\frac{1}{x})}}(\frac{1}{x^2})}{{\frac{1}{x^2}}}\]\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{1}{{(1\frac{1}{x})}}\]\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{1}{{(\frac{x1}{x})}}\]\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{x}{x1}\]Use L'Hopital's Rule again.\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{1}{1}\]\[\lim_{x \rightarrow \infty}\ln y= 1\]\[\lim_{x \rightarrow \infty}e^{\ln y}= e^1\]\[\lim_{x \rightarrow \infty}y= e^1\]\[\lim_{x \rightarrow \infty}(1\frac{1}{x})^{x}= e\]
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