## anonymous 5 years ago use l'hopital's rule to solve the limit as it approaches 0 from the right of 3/sinx - 1/x ??

1. anonymous

$\lim_{x \rightarrow 0+} 3/\sin x - 1/x$

2. anonymous

limit is 0 l'hopitals rule you'll get lim of 0/cos(0) - 0/1 so 0

3. anonymous

so 0 is also an indeternminate form?

4. anonymous

thanks by the wayyyy!!! :)

5. anonymous

no, if the limit is 0, it exists it goes to 0

6. anonymous

if it comes out to be 0 / 0 then it's indeternminate

7. anonymous

First show that limit is indeterminate, then employ l'hopital's rule lim x->0 (3x-sinx)/(xsinx)

8. anonymous

i'm not sure if cyter has it right... i'm pretty sure you deriv instead of integral for l'hopital's rule

9. anonymous

this function ends up equaling 0.. thats where i am confused bcuz why do i need to use l'hopitals rule/.

10. anonymous

lim of 0/cos(0) - 0/1 ends up zero so the limit is zero but your original equation (the question) is undefined (or indeter.) so you have to apply l'hoptials' rule

11. anonymous

It's not 0. You can't use (and don't need) l'Hopital here.

12. anonymous

polpak, can you explain your reasoning?

13. anonymous

Err wait. It's not 0, but you do need l'hopital

14. anonymous

how is my original equation (the question) undefined? 3 over sin(0) - 1 over x = 3 over 0 - 1 over 0 = 0 - 0 = 0 ??

15. myininaya

polpak when you get a chance will you come back to that kid who is trying to teach himself

16. anonymous

3/0 is not 0 it's undefined. which is why i would do l'hopitals rule but polpak says i'm wrong :x so we'll see what he says first

17. anonymous

You don't have 0 - 0, you have $$3/0 - 1/0 = \infty - \infty$$

18. anonymous

And you do need lhopital, but you need to first put over a common denominator

19. anonymous

ohhhh *faceplam* i forgot limits >< it's "as you approach" not zero thanks polpak yup itenn, polpak's right

20. anonymous

$3/sinx - 1/x = \frac{3x - sinx}{xsinx}$ Which gives a 0 over 0 and you take the derivative of top and bottom.

21. anonymous

Give that derivative a try and take the limit as the derivative aproaches 0

22. anonymous

I'll brb

23. anonymous

thankk you!!!!

24. anonymous

so far i have 3 - cos x all over 2 cos x - x sinx

25. anonymous

I agree with the numerator, but not the denominator.

26. anonymous

Using products rule $\frac{d}{dx}x(sin\ x) = x(cos\ x) + (sin\ x)$

27. anonymous

Which gives $\frac{3-cosx}{x(cos\ x) + (sin\ x)}$ Which we can then plug in 0 for x

28. anonymous

2/0 ?

29. anonymous

Yep, $$2/0^+$$

30. anonymous

So $$+\infty$$

31. anonymous

why was the first equation indeterminate again.. im sorry im not getting this:(

32. anonymous

it was similat to the 2/0+ ... it was 3/0 - 1/0.

33. anonymous

Because you had 3/0 - 1/0. Which is $$\infty - \infty$$

34. anonymous

And that can be a lot of different things. It could be 0, it could be a constant, or it could be +/- infinity.

35. anonymous

So we have to put them over a common denominator to get 0/0 or $$\infty/\infty$$ and then use l'Hopital's

36. anonymous

to see which infinity is growing faster

37. anonymous

ohkay. i did another on my own -- limit x approaches 1 [ x^3 - 1 all over x-1 ]

38. anonymous

i got 0/0 and then did derivative of top and bottom 3x^2 over 1

39. anonymous

and then i got 3. is that right?

40. anonymous

Yep, that's right.

41. anonymous

my last problem for l'hopitals rule is limit x approaches 0+ x^x ??

42. anonymous

what is the indeterminate form... if 0^0 = 0

43. anonymous

It's not 0. If you have $$\lim_{x \rightarrow c} f(x) = 0^0$$ then you have to take the $\lim_{e^x \rightarrow c} ln\ f(x)$

44. anonymous

So in this case you have $\lim_{e^x \rightarrow 0} x(ln\ x)$ Which is of the form $$0\cdot\infty$$

45. anonymous

0^0 is actually an indeterminate form so u can use l'hopital's rule

46. anonymous

and 0^0 is not equal to 0

47. anonymous

0 * inf is an indeterminate form?

48. anonymous

no

49. anonymous

polpk, I have a quick question on mine you were helping me with.. when you're done here:)

50. anonymous
51. anonymous

inf/inf, 0/0, -inf/inf, inf/-inf or 0^0, any of those are indeterminate forms

52. anonymous

$$0 * \infty$$ is indeterminate.

53. anonymous

sometimes wikipedia is unreliable

54. anonymous

the next step then would be to take the derivativve?

55. anonymous

but after futhre digging I'm convinced

56. anonymous

Was gonna say, wolfram alpha also thinks it's indeterminate: http://mathworld.wolfram.com/Indeterminate.html

57. anonymous

Which next step? I may have a different process for this than cyter.

58. anonymous

im following you polpak..

59. anonymous

I'm not sure how he was gonna use l'hopital on 0^0 since there's no fraction there, but I'm a bit rusty with this

60. anonymous

thats where i confused.. l'hopitals rule has lim f(x)/g(x) and then f'(x)/g'(x)

61. anonymous

i dont see a fraction here.

62. anonymous

Since $\lim_{e^x \rightarrow 0} x(ln\ x)$ is not nice, we have to swap our variable around a bit and say u = 1/x if x goes to 0, u goes to infinity. And that means that x = 1/u, so we have now $\lim_{e^{u} \rightarrow \infty} \frac{ln\ 1/u}{u}$ Which gives us a nice fraction with $$\infty/\infty$$ that we can take the derivative of.

63. anonymous

this one is very complicated.

64. anonymous

Yeah, a bit.

65. anonymous

i am not good with derivatices so the ln is trouble for me.. this is hard:(

66. anonymous

So $\frac{d}{du} ln\ 1/u = u*(\frac{-1}{u^2}) = \frac{-1}{u}$

67. anonymous

ohhh okay.

68. anonymous

so now we have Erg, my notation is wrong, but it's tiny if I fix it.

69. anonymous

what would change?

70. anonymous

Anyway, we have It's supposed to be e raised to the power of the limit of that whole thing, not e^u approaches infinity.

71. anonymous

So we have $e^{\lim_{u \rightarrow \infty} \frac{-1}{t}} = e^0 = 1$

72. anonymous

See, tiny.

73. anonymous

mm.. at which point?

74. anonymous

ohh it was never supposed to be e to the u

75. anonymous

Yeah, it was supposed to be e to the limit. That's it. The limit as $$x \rightarrow 0 \text{ of } x^x = 1$$

76. anonymous

so no l'hopital is needed??

77. anonymous

yes it is. I used it.

78. anonymous

This is too difficult with such a tiny font. Lemme try something else

79. anonymous

ohkay.

80. anonymous

i am confused:(

81. anonymous
82. anonymous

im there..