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

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- anonymous

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 ----------

- anonymous

infinity, I guess :) since the power will be infinity, anything to the power of infinity is infinity :)

- anonymous

or you can solve it using e and ln too

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## More answers

- anonymous

wait....that's not the case here, what's the function?

- nowhereman

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.

- anonymous

Could you please explain how to come to that step?

- nikvist

limes doesn't exist, because
\[\lim_{x\rightarrow 0-}[f(1+x)/f(1)]^{1/x}\neq\lim_{x\rightarrow 0+}[f(1+x)/f(1)]^{1/x}\]

- nikvist

wrong, my last post is not true

- nowhereman

I just put exp(ln(.)) around the function and used logarithm rules.

- anonymous

Okay, I will certainly look at it, though I have to leave again. If I can get a chance to post tonight, I will.

- anonymous

That alright?

- anonymous

Ok, al right

- nowhereman

What is it you don't understand?

- anonymous

Its ok, I understand what you said

- anonymous

I have something else to ask

- anonymous

I just wish to ask, what are the things to consider while solving problems on limit and continuity

- nowhereman

I'm sorry, but I don't know a general answer to that question. Analysis is a huge field.

- anonymous

No no, you don't have to go to that. I am in high school

- anonymous

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- anonymous

iam, I can respond now if you're around.

- anonymous

Yes I am online

- anonymous

Could you please come here
https://docs.google.com/document/d/1-_5IUkf1O4EFkBw-kDEZV1T8THk4lzrpmQEzVKtfHHo/edit?hl=en_GB#

- anonymous

\[y=\left( \frac{f(x+1)}{f(1)} \right)^{1/x} \rightarrow \log y = \log \left( \frac{f(x+1)}{f(1)} \right)^{1/x}\]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

Then\[\lim_{x->\infty}\log y=\lim_{x->\infty}\frac{\log \left( \frac{f(x+1)}{f(1)} \right)}{x}\]

- anonymous

Indeterminate, 0/0 (change that limit to 0, not infinity).

- anonymous

Use L'Hopital's rule. Then\[\lim_{x \rightarrow 0}\frac{\log \frac{f(x+1)}{f(1)}}{x}=\frac{\lim_{x \rightarrow 0}\log \frac{f(x+1)}{f(1)}}{\lim_{x \rightarrow 0}x}\]\[=\frac{\lim_{x \rightarrow 0}\frac{\left( \frac{f'(x+1)}{f(1)} \right)}{\left( \frac{f(x+1)}{f(1)} \right)}}{\lim_{x \rightarrow 0}1} =\frac{\lim_{x \rightarrow 0}f'(x+1)}{\lim_{x \rightarrow 0}f(x+1)}\]

- anonymous

No need to reply to this message. Just informing that I have received the notes

- anonymous

\[=\frac{f'(1)}{f(1)}=\frac{6}{3}=2\]

- anonymous

I have got that. I can handle the rest.

- anonymous

That is\[\lim_{x \rightarrow 0}\log y = 2 \rightarrow \lim_{x \rightarrow 0}y=e^2\]

- anonymous

Okay

- anonymous

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- anonymous

Sometimes they do.

- anonymous

This site has a good summary of limit laws:
http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html

- anonymous

Ok let me study the whole thing,
I will let you know when I need any help.
Thank you sooooooooo much

- anonymous

Is that your university?

- anonymous

No.

- anonymous

Anyhow thanks

- anonymous

np. good luck

- anonymous

I don't want to be nosy about you, as it may offend you

- anonymous

After all I am receiving something GREAT from you

- anonymous

It's okay. Just do well in STEP.

- anonymous

It will be a gift from me to you

- anonymous

I will do good, and thats a promise

- anonymous

Well, it's worth it then :)

- anonymous

Ok bye now

- anonymous

See you.

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