anonymous
  • anonymous
Pls respond
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1357625465697:dw|
anonymous
  • anonymous
ok so what do you think? :)
anonymous
  • anonymous
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.

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anonymous
  • anonymous
bamp
anonymous
  • anonymous
yup. look up l'hopital and "indeterminate" form.
anonymous
  • anonymous
tell us what you find
anonymous
  • anonymous
Dude Hero gave it away man. 1/x vanishes and then you have \[(1)^{-x}\]
anonymous
  • anonymous
@Azteck how did he give it away? . ok first get @Argos to find out what indeterminate form is -____-
anonymous
  • anonymous
i know inderminite form is 0/0 or inf/inf
anonymous
  • anonymous
and can to use lhospitals rule on those but nega exponent
anonymous
  • anonymous
Then you get this: \[\frac{ 1 }{ (1){^x} }\] \[\frac{ 1 }{ (1)^{\infty} }\] \[=\frac{ 1 }{ 1 }\]
anonymous
  • anonymous
so answer is 1?
anonymous
  • anonymous
because the back of the book sais its e
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
yeah 1 to the infinity is indeterminate so substitution will not work
shubhamsrg
  • shubhamsrg
shouldnt answer be 2 ? from binomial approximation ?
anonymous
  • anonymous
Shouldn't it be e? A more accurate answer?
shubhamsrg
  • shubhamsrg
i can see easily why e fits here.. but i dont understand why doesnt the ans come from binomial approximation
anonymous
  • anonymous
Sorry, I don't know binomial...
shubhamsrg
  • shubhamsrg
ahh,maybe i see why.. power should be a real no. for that, here, power is not a real no.
shubhamsrg
  • shubhamsrg
so, e is correct..
amoodarya
  • amoodarya
|dw:1357649297414:dw|
anonymous
  • anonymous
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:\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{\frac{1}{{(1-\frac{1}{x})}}(\frac{1}{x^2})}{{\frac{1}{x^2}}}\]\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{1}{{(1-\frac{1}{x})}}\]\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{1}{{(\frac{x-1}{x})}}\]\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{x}{x-1}\]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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