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Argos Group TitleBest ResponseYou've already chosen the best response.0
dw:1357625465697:dw
 one year ago

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

RONNCC Group TitleBest ResponseYou've already chosen the best response.0
the 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.
 one year ago

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

RONNCC Group TitleBest ResponseYou've already chosen the best response.0
tell us what you find
 one year ago

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

RONNCC Group TitleBest 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 ____
 one year ago

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

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

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

Argos Group TitleBest ResponseYou've already chosen the best response.0
so answer is 1?
 one year ago

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

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

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

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

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

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

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

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

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

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
so, e is correct..
 one year ago

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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.3
Even 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.\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{1}{1}\]\[\lim_{x \rightarrow \infty}\ln y= 1\]\[\lim_{x \rightarrow \infty}e^{\ln y}= e^1\]\[\lim_{x \rightarrow \infty}y= e^1\]\[\lim_{x \rightarrow \infty}(1\frac{1}{x})^{x}= e\]
 one year ago
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