anonymous one year ago Limit as x approaches infinity of 8x[ln(x+9) - ln(x)]

1. anonymous

How can I rewrite it as a fraction in order to apply L'Hopital's rule?

2. hartnn

you can bring x in the denominator and write it as 1/x

3. hartnn

|dw:1435026931030:dw|

4. anonymous

Right but then plugging in infinity I would get infinity/0 right? And thats not defined

5. hartnn

you wanted to apply LH right ? so first differentiate numerator and denominator separately later think about plugging it, after some simplification

6. anonymous

Thats true but in order to use LH we have to first make sure that we have an indeterminate form and oo/0 isn't it

7. hartnn

right, did you try the substitution x =1/y ? when x-> infinity, y->0

8. SolomonZelman

$$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\left(\ln(x+9)-\ln(x)\right)}$$ $$\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\left(\ln\frac{x+9}{x}\right)}$$ $$\large\color{slate}{\displaystyle\left(\ln\left(\lim_{x \rightarrow ~\infty }\frac{x+9}{x}\right)\right)}$$

9. SolomonZelman

so you are taking the ln of that limit

10. SolomonZelman

0

11. SolomonZelman

oh 8x !!1soprry

12. anonymous

Yes thats right! Oh but where's the 8x?

13. SolomonZelman

14. anonymous

And so is that also one of the limit properties? Exchanging the places of ln and limit?

15. SolomonZelman

yes limit properties allow thatr

16. anonymous

Alright so using the same idea where would the 8x go?

17. hartnn

its not exchanging... Limit can be dragged inside of only those functions which are continuous

18. hartnn

can you give the substitution one shot? the substitution x =1/y ? when x-> infinity, y->0

19. anonymous

Ohh now that makes sense! I have to review those properties

20. anonymous

How would I apply that method @hartnn

21. hartnn

plug in x= 1/y so y = 1/x -> 1/infinity->0 8x becomes 8/y ln (x+9) becomes ln (1/y + 9) ln x becomes ln (1/y) or

22. anonymous

Okay I understand the steps.. just not clear on the concept, why are we saying that x=1/y?

23. SolomonZelman

limit change is fine

24. SolomonZelman

if not the only...

25. hartnn

its a substitution, it may/may not make the solution easier/faster to find... just a try

26. anonymous

Ohh ok

27. hartnn

can you try to combine 2 logs? $$\ln (1/y+9) - \ln (1/y)$$

28. hartnn

ln A - ln B = ln A/B

29. hartnn

the limit really becomes simple after that...you'll easily be able to apply LH rule

30. anonymous

That would be ln[ (1/y +9)/(1/y)]

31. hartnn

yep, simplify that further

32. hartnn

|dw:1435028541306:dw|

33. anonymous

So it simplifies to ... yah exactly haha! I was just about to type that in

34. hartnn

good! now is your new limit in the form of 0/0? :)

35. anonymous

hey umm sorry to interrupt but could u help me pleeaassee!

36. anonymous

its urgent

37. anonymous

But we no longer have a fraction, do we?

38. hartnn

remember the 8x became 8/y?

39. anonymous

Oh yeah!!

40. anonymous

so the we have (8ln(1+9y))/y which is equal to 0/0 ?

41. anonymous

then*

42. hartnn

yes :3

43. UsukiDoll

if it's 0/0 that means you can use LH rule again

44. anonymous

Alrighty! Yeah exactly!

45. UsukiDoll

When you don't have the 0/0 form anymore, you can't use the rule.

46. hartnn

yaay :D

47. anonymous

So now to differentiate, do I use implicit diff ?

48. hartnn

Implicit? you mean chain rule for numerator?

49. anonymous

Or do i just change the y's to 1/x again?

50. UsukiDoll

$\frac{(8\ln(1+9y))}{y}$ this?

51. hartnn

thats the beauty of substitution, you don't need to change back :)

52. hartnn

53. anonymous

Im just thinking of the y's? do i diff with respect to y then?

54. hartnn

yes, diff. w.r.t y only

55. anonymous

Ohh ok that makes sense!

56. anonymous

product rule on the top?

57. hartnn

there are no productions on the top? chain rule... its like a function inside a function linear function inside log function

58. anonymous

theres also an 8y tho

59. anonymous

oh no the y's on the bottom haha sorry!

60. hartnn

8y ?? there is only 8 in the numerator, right ? 8 is a constant and can be brought out.. lol ok

61. anonymous

okay so I'm hoping i got it right in the end... is it 8/1+9y

62. UsukiDoll

@hartnn constants don't have derivatives... it goes to 0 Ex. the derivative of 6 is 0

63. hartnn

not exactly $$[\ln (1+9y)]' = (1/(1+9y))(1+9y)'$$ chain rule you missed differentiating 1+9y

64. hartnn

thats right Usuki, here the constant 8 is multiplied with log function its not alone thats why we can bring it out of limit

65. anonymous

8*9 / 1+9y ?

66. hartnn

yep now you can just plug in y= 0 :)

67. anonymous

Ahhh so that leaves us with 8*9 = 72? :O

68. hartnn

$$\huge \checkmark$$

69. anonymous

Hooray!!! Thats correct! :D

70. hartnn

congrats! and welcome ^_^

71. anonymous

Thanks so much!! I was getting totally hopeless on this one xD

72. anonymous

Thanks to everyone who helped out! :)