A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

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.\[\lim_{x \rightarrow \infty}\ln y= \lim_{x \rightarrow \infty}\frac{1}{1}\]\[\lim_{x \rightarrow \infty}\ln y= 1\]\[\lim_{x \rightarrow \infty}e^{\ln y}= e^1\]\[\lim_{x \rightarrow \infty}y= e^1\]\[\lim_{x \rightarrow \infty}(1\frac{1}{x})^{x}= e\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.