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anonymous
 5 years ago
Let f=R>R be such that f(1)=3 and f '(1)=6 , then lim(x tends to zero) [(f(1+x)/f(1)]^(1/x) equals to 
anonymous
 5 years ago
Let f=R>R be such that f(1)=3 and f '(1)=6 , then lim(x tends to zero) [(f(1+x)/f(1)]^(1/x) equals to 

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let f=R>R be such that f(1)=3 and f '(1)=6 , then \[\lim_{x \rightarrow 0}[(f(1+x)/f(1)]^{1/x}\] equals to 

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0infinity, I guess :) since the power will be infinity, anything to the power of infinity is infinity :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or you can solve it using e and ln too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait....that's not the case here, what's the function?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0you don't need the function Just write it as \[\exp\left(\frac {\ln \frac{ f(1+x)}{f(1)}}{x}\right)\] and you can solve it with l'hopital.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Could you please explain how to come to that step?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0limes doesn't exist, because \[\lim_{x\rightarrow 0}[f(1+x)/f(1)]^{1/x}\neq\lim_{x\rightarrow 0+}[f(1+x)/f(1)]^{1/x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wrong, my last post is not true

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0I just put exp(ln(.)) around the function and used logarithm rules.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, I will certainly look at it, though I have to leave again. If I can get a chance to post tonight, I will.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0What is it you don't understand?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Its ok, I understand what you said

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have something else to ask

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I just wish to ask, what are the things to consider while solving problems on limit and continuity

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0I'm sorry, but I don't know a general answer to that question. Analysis is a huge field.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No no, you don't have to go to that. I am in high school

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0AYCC is a homework tutoring marketplace. If you're a student seeking smart solutions to your difficult homework assignments or someone who’d love to pass on your pearls of wisdom for a specific topic or a question and earn additional income, join www.aceyourcollegeclasses.com

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0iam, I can respond now if you're around.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Could you please come here https://docs.google.com/document/d/1_5IUkf1O4EFkBwkDEZV1T8THk4lzrpmQEzVKtfHHo/edit?hl=en_GB#

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=\left( \frac{f(x+1)}{f(1)} \right)^{1/x} \rightarrow \log y = \log \left( \frac{f(x+1)}{f(1)} \right)^{1/x}\]That is\[\log y =\frac{1}{x}\log \left( \frac{f(x+1)}{f(1)} \right)=\frac{\log \left( \frac{f(x+1)}{f(1)} \right)}{x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then\[\lim_{x>\infty}\log y=\lim_{x>\infty}\frac{\log \left( \frac{f(x+1)}{f(1)} \right)}{x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Indeterminate, 0/0 (change that limit to 0, not infinity).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use L'Hopital's rule. Then\[\lim_{x \rightarrow 0}\frac{\log \frac{f(x+1)}{f(1)}}{x}=\frac{\lim_{x \rightarrow 0}\log \frac{f(x+1)}{f(1)}}{\lim_{x \rightarrow 0}x}\]\[=\frac{\lim_{x \rightarrow 0}\frac{\left( \frac{f'(x+1)}{f(1)} \right)}{\left( \frac{f(x+1)}{f(1)} \right)}}{\lim_{x \rightarrow 0}1} =\frac{\lim_{x \rightarrow 0}f'(x+1)}{\lim_{x \rightarrow 0}f(x+1)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No need to reply to this message. Just informing that I have received the notes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{f'(1)}{f(1)}=\frac{6}{3}=2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have got that. I can handle the rest.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That is\[\lim_{x \rightarrow 0}\log y = 2 \rightarrow \lim_{x \rightarrow 0}y=e^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are really a good person. In this world no one does so much for any other without any return

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This site has a good summary of limit laws: http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok let me study the whole thing, I will let you know when I need any help. Thank you sooooooooo much

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that your university?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't want to be nosy about you, as it may offend you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0After all I am receiving something GREAT from you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's okay. Just do well in STEP.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It will be a gift from me to you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I will do good, and thats a promise

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, it's worth it then :)
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