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

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

Azteck
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0i know inderminite form is 0/0 or inf/inf

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

Azteck
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0because the back of the book sais its e

Argos
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0yeah 1 to the infinity is indeterminate so substitution will not work

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

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

shubhamsrg
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.3Sorry, I don't know binomial...

shubhamsrg
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1357649297414:dw

RolyPoly
 2 years 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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