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anonymous

  • one year ago

Limit as x approaches infinity of 8x[ln(x+9) - ln(x)]

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  1. anonymous
    • one year ago
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    How can I rewrite it as a fraction in order to apply L'Hopital's rule?

  2. hartnn
    • one year ago
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    you can bring x in the denominator and write it as 1/x

  3. hartnn
    • one year ago
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    |dw:1435026931030:dw|

  4. anonymous
    • one year ago
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    Right but then plugging in infinity I would get infinity/0 right? And thats not defined

  5. hartnn
    • one year ago
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    you wanted to apply LH right ? so first differentiate numerator and denominator separately later think about plugging it, after some simplification

  6. anonymous
    • one year ago
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    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
    • one year ago
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    right, did you try the substitution x =1/y ? when x-> infinity, y->0

  8. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\left(\ln(x+9)-\ln(x)\right)}\) \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\left(\ln\frac{x+9}{x}\right)}\) \(\large\color{slate}{\displaystyle\left(\ln\left(\lim_{x \rightarrow ~\infty }\frac{x+9}{x}\right)\right)}\)

  9. SolomonZelman
    • one year ago
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    so you are taking the ln of that limit

  10. SolomonZelman
    • one year ago
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    0

  11. SolomonZelman
    • one year ago
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    oh 8x !!1soprry

  12. anonymous
    • one year ago
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    Yes thats right! Oh but where's the 8x?

  13. SolomonZelman
    • one year ago
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    yes yes, my bad

  14. anonymous
    • one year ago
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    And so is that also one of the limit properties? Exchanging the places of ln and limit?

  15. SolomonZelman
    • one year ago
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    yes limit properties allow thatr

  16. anonymous
    • one year ago
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    Alright so using the same idea where would the 8x go?

  17. hartnn
    • one year ago
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    its not exchanging... Limit can be dragged inside of only those functions which are continuous

  18. hartnn
    • one year ago
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    can you give the substitution one shot? the substitution x =1/y ? when x-> infinity, y->0

  19. anonymous
    • one year ago
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    Ohh now that makes sense! I have to review those properties

  20. anonymous
    • one year ago
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    How would I apply that method @hartnn

  21. hartnn
    • one year ago
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    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
    • one year ago
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    Okay I understand the steps.. just not clear on the concept, why are we saying that x=1/y?

  23. SolomonZelman
    • one year ago
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    limit change is fine

  24. SolomonZelman
    • one year ago
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    if not the only...

  25. hartnn
    • one year ago
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    its a substitution, it may/may not make the solution easier/faster to find... just a try

  26. anonymous
    • one year ago
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    Ohh ok

  27. hartnn
    • one year ago
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    can you try to combine 2 logs? \(\ln (1/y+9) - \ln (1/y)\)

  28. hartnn
    • one year ago
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    ln A - ln B = ln A/B

  29. hartnn
    • one year ago
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    the limit really becomes simple after that...you'll easily be able to apply LH rule

  30. anonymous
    • one year ago
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    That would be ln[ (1/y +9)/(1/y)]

  31. hartnn
    • one year ago
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    yep, simplify that further

  32. hartnn
    • one year ago
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    |dw:1435028541306:dw|

  33. anonymous
    • one year ago
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    So it simplifies to ... yah exactly haha! I was just about to type that in

  34. hartnn
    • one year ago
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    good! now is your new limit in the form of 0/0? :)

  35. anonymous
    • one year ago
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    hey umm sorry to interrupt but could u help me pleeaassee!

  36. anonymous
    • one year ago
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    its urgent

  37. anonymous
    • one year ago
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    But we no longer have a fraction, do we?

  38. hartnn
    • one year ago
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    remember the 8x became 8/y?

  39. anonymous
    • one year ago
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    Oh yeah!!

  40. anonymous
    • one year ago
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    so the we have (8ln(1+9y))/y which is equal to 0/0 ?

  41. anonymous
    • one year ago
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    then*

  42. hartnn
    • one year ago
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    yes :3

  43. UsukiDoll
    • one year ago
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    if it's 0/0 that means you can use LH rule again

  44. anonymous
    • one year ago
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    Alrighty! Yeah exactly!

  45. UsukiDoll
    • one year ago
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    When you don't have the 0/0 form anymore, you can't use the rule.

  46. hartnn
    • one year ago
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    yaay :D

  47. anonymous
    • one year ago
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    So now to differentiate, do I use implicit diff ?

  48. hartnn
    • one year ago
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    Implicit? you mean chain rule for numerator?

  49. anonymous
    • one year ago
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    Or do i just change the y's to 1/x again?

  50. UsukiDoll
    • one year ago
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    \[\frac{(8\ln(1+9y))}{y}\] this?

  51. hartnn
    • one year ago
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    thats the beauty of substitution, you don't need to change back :)

  52. hartnn
    • one year ago
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    whatever answer you get for that limit, will be the answer for your original limit

  53. anonymous
    • one year ago
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    Im just thinking of the y's? do i diff with respect to y then?

  54. hartnn
    • one year ago
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    yes, diff. w.r.t y only

  55. anonymous
    • one year ago
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    Ohh ok that makes sense!

  56. anonymous
    • one year ago
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    product rule on the top?

  57. hartnn
    • one year ago
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    there are no productions on the top? chain rule... its like a function inside a function linear function inside log function

  58. anonymous
    • one year ago
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    theres also an 8y tho

  59. anonymous
    • one year ago
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    oh no the y's on the bottom haha sorry!

  60. hartnn
    • one year ago
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    8y ?? there is only 8 in the numerator, right ? 8 is a constant and can be brought out.. lol ok

  61. anonymous
    • one year ago
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    okay so I'm hoping i got it right in the end... is it 8/1+9y

  62. UsukiDoll
    • one year ago
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    @hartnn constants don't have derivatives... it goes to 0 Ex. the derivative of 6 is 0

  63. hartnn
    • one year ago
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    not exactly \([\ln (1+9y)]' = (1/(1+9y))(1+9y)'\) chain rule you missed differentiating 1+9y

  64. hartnn
    • one year ago
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    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
    • one year ago
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    8*9 / 1+9y ?

  66. hartnn
    • one year ago
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    yep now you can just plug in y= 0 :)

  67. anonymous
    • one year ago
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    Ahhh so that leaves us with 8*9 = 72? :O

  68. hartnn
    • one year ago
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    \(\huge \checkmark \)

  69. anonymous
    • one year ago
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    Hooray!!! Thats correct! :D

  70. hartnn
    • one year ago
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    congrats! and welcome ^_^

  71. anonymous
    • one year ago
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    Thanks so much!! I was getting totally hopeless on this one xD

  72. anonymous
    • one year ago
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    Thanks to everyone who helped out! :)

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