## 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 ----------

1. 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 ----------

2. anonymous

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

3. anonymous

or you can solve it using e and ln too

4. anonymous

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

5. 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.

6. anonymous

Could you please explain how to come to that step?

7. anonymous

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

8. anonymous

wrong, my last post is not true

9. nowhereman

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

10. 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.

11. anonymous

That alright?

12. anonymous

Ok, al right

13. nowhereman

What is it you don't understand?

14. anonymous

Its ok, I understand what you said

15. anonymous

I have something else to ask

16. anonymous

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

17. nowhereman

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

18. anonymous

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

19. anonymous

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20. anonymous

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

21. anonymous

Yes I am online

22. anonymous

23. 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$\log y =\frac{1}{x}\log \left( \frac{f(x+1)}{f(1)} \right)=\frac{\log \left( \frac{f(x+1)}{f(1)} \right)}{x}$

24. anonymous

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

25. anonymous

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

26. 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)}$

27. anonymous

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

28. anonymous

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

29. anonymous

I have got that. I can handle the rest.

30. anonymous

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

31. anonymous

Okay

32. anonymous

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33. anonymous

Sometimes they do.

34. anonymous

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

35. anonymous

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

36. anonymous

37. anonymous

No.

38. anonymous

Anyhow thanks

39. anonymous

np. good luck

40. anonymous

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41. anonymous

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42. anonymous

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43. anonymous

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44. anonymous

I will do good, and thats a promise

45. anonymous

Well, it's worth it then :)

46. anonymous

Ok bye now

47. anonymous

See you.