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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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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Let 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 ----------
infinity, I guess :) since the power will be infinity, anything to the power of infinity is infinity :)
or you can solve it using e and ln too

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wait....that's not the case here, what's the function?
you 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.
Could you please explain how to come to that step?
limes doesn't exist, because \[\lim_{x\rightarrow 0-}[f(1+x)/f(1)]^{1/x}\neq\lim_{x\rightarrow 0+}[f(1+x)/f(1)]^{1/x}\]
wrong, my last post is not true
I just put exp(ln(.)) around the function and used logarithm rules.
Okay, I will certainly look at it, though I have to leave again. If I can get a chance to post tonight, I will.
That alright?
Ok, al right
What is it you don't understand?
Its ok, I understand what you said
I have something else to ask
I just wish to ask, what are the things to consider while solving problems on limit and continuity
I'm sorry, but I don't know a general answer to that question. Analysis is a huge field.
No no, you don't have to go to that. I am in high school
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iam, I can respond now if you're around.
Yes I am online
Could you please come here https://docs.google.com/document/d/1-_5IUkf1O4EFkBw-kDEZV1T8THk4lzrpmQEzVKtfHHo/edit?hl=en_GB#
\[y=\left( \frac{f(x+1)}{f(1)} \right)^{1/x} \rightarrow \log y = \log \left( \frac{f(x+1)}{f(1)} \right)^{1/x}\]That is\[\log y =\frac{1}{x}\log \left( \frac{f(x+1)}{f(1)} \right)=\frac{\log \left( \frac{f(x+1)}{f(1)} \right)}{x}\]
Then\[\lim_{x->\infty}\log y=\lim_{x->\infty}\frac{\log \left( \frac{f(x+1)}{f(1)} \right)}{x}\]
Indeterminate, 0/0 (change that limit to 0, not infinity).
Use L'Hopital's rule. Then\[\lim_{x \rightarrow 0}\frac{\log \frac{f(x+1)}{f(1)}}{x}=\frac{\lim_{x \rightarrow 0}\log \frac{f(x+1)}{f(1)}}{\lim_{x \rightarrow 0}x}\]\[=\frac{\lim_{x \rightarrow 0}\frac{\left( \frac{f'(x+1)}{f(1)} \right)}{\left( \frac{f(x+1)}{f(1)} \right)}}{\lim_{x \rightarrow 0}1} =\frac{\lim_{x \rightarrow 0}f'(x+1)}{\lim_{x \rightarrow 0}f(x+1)}\]
No need to reply to this message. Just informing that I have received the notes
\[=\frac{f'(1)}{f(1)}=\frac{6}{3}=2\]
I have got that. I can handle the rest.
That is\[\lim_{x \rightarrow 0}\log y = 2 \rightarrow \lim_{x \rightarrow 0}y=e^2\]
Okay
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Sometimes they do.
This site has a good summary of limit laws: http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html
Ok let me study the whole thing, I will let you know when I need any help. Thank you sooooooooo much
Is that your university?
No.
Anyhow thanks
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See you.

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