anonymous
  • anonymous
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 ----------
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • anonymous
infinity, I guess :) since the power will be infinity, anything to the power of infinity is infinity :)
anonymous
  • anonymous
or you can solve it using e and ln too

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
wait....that's not the case here, what's the function?
nowhereman
  • 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
  • anonymous
Could you please explain how to come to that step?
nikvist
  • 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
  • nikvist
wrong, my last post is not true
nowhereman
  • nowhereman
I just put exp(ln(.)) around the function and used logarithm rules.
anonymous
  • 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
  • anonymous
That alright?
anonymous
  • anonymous
Ok, al right
nowhereman
  • nowhereman
What is it you don't understand?
anonymous
  • anonymous
Its ok, I understand what you said
anonymous
  • anonymous
I have something else to ask
anonymous
  • anonymous
I just wish to ask, what are the things to consider while solving problems on limit and continuity
nowhereman
  • nowhereman
I'm sorry, but I don't know a general answer to that question. Analysis is a huge field.
anonymous
  • anonymous
No no, you don't have to go to that. I am in high school
anonymous
  • anonymous
AYCC 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
  • anonymous
iam, I can respond now if you're around.
anonymous
  • anonymous
Yes I am online
anonymous
  • anonymous
Could you please come here https://docs.google.com/document/d/1-_5IUkf1O4EFkBw-kDEZV1T8THk4lzrpmQEzVKtfHHo/edit?hl=en_GB#
anonymous
  • 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
  • anonymous
Then\[\lim_{x->\infty}\log y=\lim_{x->\infty}\frac{\log \left( \frac{f(x+1)}{f(1)} \right)}{x}\]
anonymous
  • anonymous
Indeterminate, 0/0 (change that limit to 0, not infinity).
anonymous
  • 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
  • anonymous
No need to reply to this message. Just informing that I have received the notes
anonymous
  • anonymous
\[=\frac{f'(1)}{f(1)}=\frac{6}{3}=2\]
anonymous
  • anonymous
I have got that. I can handle the rest.
anonymous
  • anonymous
That is\[\lim_{x \rightarrow 0}\log y = 2 \rightarrow \lim_{x \rightarrow 0}y=e^2\]
anonymous
  • anonymous
Okay
anonymous
  • anonymous
You are really a good person. In this world no one does so much for any other without any return
anonymous
  • anonymous
Sometimes they do.
anonymous
  • 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
  • anonymous
Ok let me study the whole thing, I will let you know when I need any help. Thank you sooooooooo much
anonymous
  • anonymous
Is that your university?
anonymous
  • anonymous
No.
anonymous
  • anonymous
Anyhow thanks
anonymous
  • anonymous
np. good luck
anonymous
  • anonymous
I don't want to be nosy about you, as it may offend you
anonymous
  • anonymous
After all I am receiving something GREAT from you
anonymous
  • anonymous
It's okay. Just do well in STEP.
anonymous
  • anonymous
It will be a gift from me to you
anonymous
  • anonymous
I will do good, and thats a promise
anonymous
  • anonymous
Well, it's worth it then :)
anonymous
  • anonymous
Ok bye now
anonymous
  • anonymous
See you.

Looking for something else?

Not the answer you are looking for? Search for more explanations.