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

1. GChamon Group Title

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

2. slaaibak Group Title

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

3. GChamon Group Title

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 Group Title

I'll quickly get a pen and try

5. slaaibak Group Title

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

6. GChamon Group Title

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

7. satellite73 Group Title

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

8. GChamon Group Title

yeah

9. slaaibak Group Title

you mean the limit to zero?

10. satellite73 Group Title

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

11. GChamon Group Title

as x tends to 0

12. satellite73 Group Title

oops ok lets try again

13. satellite73 Group Title

and no l'hopital??

14. slaaibak Group Title

lol apparently not hey...

15. TuringTest Group Title

that is cruel

16. GChamon Group Title

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

17. satellite73 Group Title

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

18. satellite73 Group Title

then take the limit as z goes to infinity

19. GChamon Group Title

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

20. slaaibak Group Title

I think I got it. Will explain now

21. satellite73 Group Title

oh lord i am in idiot!

22. TuringTest Group Title

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

23. TuringTest Group Title

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

24. satellite73 Group Title

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

25. satellite73 Group Title

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

26. slaaibak Group Title

That will only result in another indeterminate form

27. satellite73 Group Title

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

28. slaaibak Group Title

kk I got it, will post method now:

29. TuringTest Group Title

go slaiibak!

30. satellite73 Group Title

can't wait! no derivatives right?

31. Kreshnik Group Title

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 Group Title

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

33. slaaibak Group Title

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

34. Kreshnik Group Title

@satellite73 LOL

35. GChamon Group Title

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

36. GChamon Group Title

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

37. TuringTest Group Title

I doubt it

38. slaaibak Group Title

$\exp(xln(2x+1)) = \exp({x^2 \ln(2x + 1) \over x})$ 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 Group Title

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

40. slaaibak Group Title

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

41. TuringTest Group Title

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

42. GChamon Group Title

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 Group Title

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 Group Title

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 Group Title

wow that was pretty snazzy!

46. GChamon Group Title

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

47. slaaibak Group Title

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 Group Title

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