GChamon 3 years ago want some help with this Limit lim(x->0) (1+1/(2*x))^x at least a guidance! thanks!

1. GChamon

$\lim_{x \rightarrow 0} (1+1/2x)^x$

2. slaaibak

rewrite it as: e^(ln(1 + 1/2x)x) then manipulate it into l'hopital

3. GChamon

thing is, we still haven't got to l'hôspital even though I know what ur talkign about. Teacher is not gonna consider any answer using l'hôspital... a real setback to these exercises. anyway to solve this without this tool?

4. slaaibak

I'll quickly get a pen and try

5. slaaibak

Just making sure, it's 2x in the denominator, right?

6. GChamon

yeah is 1 over 2*x. this is why it is bugging me cuz i can't get rid of the indetermination

7. satellite73

is it $\lim_{x\to \infty}(1+\frac{1}{2x})^x$?

8. GChamon

yeah

9. slaaibak

you mean the limit to zero?

10. satellite73

ok think of it as $(1+\frac{1}{2}\times \frac{1}{x})^x$

11. GChamon

as x tends to 0

12. satellite73

oops ok lets try again

13. satellite73

and no l'hopital??

14. slaaibak

lol apparently not hey...

15. TuringTest

that is cruel

16. GChamon

ikr... such a wonderful tool. is like deriving using the concept of rate of change

17. satellite73

ok maybe replace $$x$$ by $$\frac{1}{z}$$

18. satellite73

then take the limit as z goes to infinity

19. GChamon

ok, so using l'hôspital, is the answer 1?

20. slaaibak

I think I got it. Will explain now

21. satellite73

oh lord i am in idiot!

22. TuringTest

but then we get$\exp\left(\lim_{z\to \infty}{\ln(1+\frac z2)\over z}\right)$

23. TuringTest

if you're an idiot satellite I hate to think what that makes me !

24. satellite73

$e^{x\ln(1+\frac{1}{2x}}=e^{x\ln(\frac{2x+1}{2x})}$

25. satellite73

tak the limit as x goes to zero oh wait nope not so snappy, ok i am not such an idiot

26. slaaibak

That will only result in another indeterminate form

27. satellite73

right i see that. for a moment i thought it was $$\ln(1)=0$$ but it is not damn!

28. slaaibak

kk I got it, will post method now:

29. TuringTest

go slaiibak!

30. satellite73

can't wait! no derivatives right?

31. Kreshnik

You're supposed to know this rule: $\LARGE \lim_{x\to \infty}\left(1+\frac{1}{x}\right)^x=e$ $\Large \lim_{x\to 0} \left(1+\frac{1}{2x}\right)^x$ $\Large \lim_{x\to 0} \left[\left(1+\frac{1}{2x}\right)^{\frac{2x}{1}\cdot \frac{1}{2x }}\right]^x$ $\Large \lim_{x\to 0} \left[\left(1+\frac{1}{2x}\right)^{2x} \right]^{\frac{x}{2x}}$ $\LARGE \left(e\right)^{\frac12}=\sqrt e$ AAhhh.. LOOK WHAT I'VE DONE !! NOW I SEE THAT IT'S NOT INFINITE BUT IS "0" AUFFF... ( at least I'll post it :P ) hahahaah....

32. satellite73

lol that is what i got the first try, but i thought it was as x goes to infinity too !

33. slaaibak

It does have a derivative in, but it's not L"Hospital. Or it's arguable, but will post quickly..

34. Kreshnik

@satellite73 LOL

35. GChamon

good lord... i'll have to spend these two days studying my retriceoff. have test on wednesday

36. GChamon

whould it help to work the problem backwards? (from the answer)

37. TuringTest

I doubt it

38. slaaibak

$\exp(xln(2x+1)) = \exp({x^2 \ln(2x + 1) \over x})$ now using this: $\lim_{x \rightarrow a} {f(x) - f(a) \over x - a} = f'(a)$ setting f(x) = x^2ln(2x + 1) we get $\exp( \lim_{x \rightarrow 0} {x^2\ln(2x + 1) - (0)^2 \ln(2(0) + 1)\over x - 0} ) = \exp(f'(0))$

39. TuringTest

we can all get the answer with l'hospital, but you said $$no$$ $$l'hospital$$ so we are doing loops for you

40. slaaibak

f'(x) = 2x(ln(2x+1) + x^2 / (2x + 1) f'(0) = 0 therefore e^0 = 1. So the limit = 1

41. TuringTest

well, that's a proof all right I guess since it doesn't technically use l'Hospital it's okay nice one!

42. GChamon

would anyone bother to talk me a little bit through the proof? exp(xln(2x+1))=exp(x2ln(2x+1)x) how do u get to this in the first place? i know it sounds dumb but i am new to calculus

43. slaaibak

It's just a little bit of algebraic manipulation to force it into the f'(a) form. Check if you simplify, you will get back to the original thing.

44. GChamon

alright! gonna put some effort into learning more about functions since it is somehow not so clear to me how to manipulate the original expression. nevertheless, surely u guys helped a lot! thanks!!

45. satellite73

wow that was pretty snazzy!

46. GChamon

haha, got my english from a brit school (I am brazillian)

47. slaaibak

GChamon, as you do more problems and exercises, you'll start to see it much clearer. Since you're only in the start of calculus now, you'll learn so much more and these things will become more intuitive. Good luck with the test!

48. GChamon

thanks! that cheers me little to study! they all told me engineering is not easy. starting to see they were all right