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anonymous

  • one year ago

Can you please show me how to solve this?I would really appreciete it! (Limits and series) http://tinypic.com/r/dy3x4m/8

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  1. Jhannybean
    • one year ago
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    \[\lim_{x\rightarrow -1} \frac{x^2+3x+2}{2x^2-8} \\ \lim_{x\rightarrow -2} \frac{x^2+3x+2}{2x^2-8}\] \[\lim_{x\rightarrow 1} \frac{\sqrt{1+3x}-2}{x-1}\] \[\lim_{(x~,y)\rightarrow (0,0)}\frac{x^2}{\sqrt{x^2+y^2}} \\ \lim_{(x~,~y)\rightarrow (0,0)} \frac{xy}{\sqrt{x^2+y^2}}\]

  2. Jhannybean
    • one year ago
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    Ready for a race? @zzr0ck3r haha.

  3. zzr0ck3r
    • one year ago
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    Heck no. I hate calculus :)

  4. Jhannybean
    • one year ago
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    Okay in limit #1, if you directly plug in -1 for all the x's, what do you get, @xfire30 ?

  5. Jhannybean
    • one year ago
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    We can apply L'Hospital's rule (if you're allowed to) in limit #2. This will become \(\sf \dfrac{2x+3}{4x}\) and plug in -2

  6. anonymous
    • one year ago
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    6/-6 i get in limit #!?

  7. anonymous
    • one year ago
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    is it 6/-6?

  8. Jhannybean
    • one year ago
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    For limit #1?

  9. anonymous
    • one year ago
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    yes

  10. Jhannybean
    • one year ago
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    no. \((-1)^2+3(-1)+2 \ne 6\) but \(2(-1)^2-8 =-6\)

  11. anonymous
    • one year ago
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    0/-6

  12. Jhannybean
    • one year ago
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    Yes and what does that =.. plugging it into a calculator would be sooo simple.

  13. anonymous
    • one year ago
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    0

  14. Jhannybean
    • one year ago
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    good. limit #2, are you allowed to apply LH rule?

  15. anonymous
    • one year ago
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    I am allowed

  16. Jhannybean
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @Jhannybean We can apply L'Hospital's rule (if you're allowed to) in limit #2. This will become \(\sf \dfrac{2x+3}{4x}\) and plug in -2 \(\color{#0cbb34}{\text{End of Quote}}\)

  17. anonymous
    • one year ago
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    I dotn get it

  18. anonymous
    • one year ago
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    How did he got (2x+3)/(4x)?

  19. anonymous
    • one year ago
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    you get*

  20. anonymous
    • one year ago
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    Ohhh,I get it now

  21. Jhannybean
    • one year ago
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    When you apply L'Hospital's rule, you tage the derivative of the function in the numerator and the denominator. Ok cool.

  22. anonymous
    • one year ago
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    Derivated each term

  23. anonymous
    • one year ago
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    The first 2 I understood,thank you so far :D

  24. Jhannybean
    • one year ago
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    Yep, you derivitated each term. woot, np.

  25. anonymous
    • one year ago
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    And now I plug in x=1?

  26. anonymous
    • one year ago
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    How do I solve it?

  27. anonymous
    • one year ago
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    Looks the same to me,except that little "+" and"-"

  28. Astrophysics
    • one year ago
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    It's sort of like the squeeze theorem if they both approach the same number/ exists, then the two sided limit will also approach the same number as the one sided limits approach.

  29. Astrophysics
    • one year ago
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    \[\lim_{x \rightarrow c} f(x) = L\] then the limit approaching from left and right equal the same number L pretty much

  30. Jhannybean
    • one year ago
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    I know it doesn't exist, but when I solve it it seems like it freaking does. Lol. Wth :(

  31. Astrophysics
    • one year ago
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    It should exist

  32. Astrophysics
    • one year ago
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    Mess around with it, apply L'Hopital's rule

  33. Jhannybean
    • one year ago
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    Ok so no left / right hand limits.

  34. Jhannybean
    • one year ago
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    wolfram gives 3/4? http://www.wolframalpha.com/input/?i=lim+x-%3E+1+%28%28%5Csqrt%281%2B3x%29-2%29%2F%28x-1%29%29

  35. Astrophysics
    • one year ago
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    No there is left and right hand limits, and they equal, that's why it exists :P

  36. Astrophysics
    • one year ago
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    Sounds about right

  37. Jhannybean
    • one year ago
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    oh. dur... I forgot to apply chain rule.

  38. Jhannybean
    • one year ago
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    \[\sf \frac{d}{dx} (\sqrt{1+3x}-2) =\frac{1}{2}(1+3x)^{- 1/2}(3)\] so you aplly LH rule I guess haha.

  39. Jhannybean
    • one year ago
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    \[\sf \lim_{x\rightarrow 1} \frac{\sqrt{1+3x}-2}{x-1}\]Applying LH rule, you get: \[\lim_{x\rightarrow 1} \frac{\frac{1}{2}(1+3x)^{-1/2}(3)}{1}\]Now you just plug in 1 and you'll get your limit! Sorry about that earlier.

  40. anonymous
    • one year ago
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    Dont worry,just give me a second to try to understand what you wrote :D

  41. anonymous
    • one year ago
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    This one is hard

  42. anonymous
    • one year ago
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    You just derivated each term,right?

  43. Jhannybean
    • one year ago
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    yep

  44. anonymous
    • one year ago
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    What's "d" and "dx"?

  45. Jhannybean
    • one year ago
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    that's how you write the derivative, lol

  46. Jhannybean
    • one year ago
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    \(\sf \frac{d}{dx}\) means im taking the derivative, \(\sf d\), with respect to x. \(\sf dx\)

  47. anonymous
    • one year ago
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    (1+3x)^(−1/2)

  48. anonymous
    • one year ago
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    how do i solve this part?

  49. Jhannybean
    • one year ago
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    Yeah and don't forget the chain rule

  50. anonymous
    • one year ago
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    4^(-1/2) how much is it?

  51. Jhannybean
    • one year ago
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    No, no. Are you familiar with derivatives?

  52. anonymous
    • one year ago
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    \[4^{\frac{ -1 }{ 2 }}\]

  53. anonymous
    • one year ago
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    I just plugged X=1 in the result you got after derivating

  54. Jhannybean
    • one year ago
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    Yeah, I just plugged it all into my calculator.

  55. Jhannybean
    • one year ago
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    (1/2)* (1+ 3(1))^(-1/2) * (3)

  56. anonymous
    • one year ago
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    (1+ 3(1))^(-1/2) this part I dont know how to calculate

  57. Jhannybean
    • one year ago
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    You can also think of it as \(\dfrac{3}{2(1+3(1))^{1/2}}\)

  58. Jhannybean
    • one year ago
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    You make a negative power positive by sending it to the nether realms, aka denominator

  59. anonymous
    • one year ago
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    the fraction is the problem I dont know to solve as power

  60. anonymous
    • one year ago
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    something^1/2

  61. anonymous
    • one year ago
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    how is it done?

  62. Jhannybean
    • one year ago
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    the 1/2 power is the square root.

  63. Jhannybean
    • one year ago
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    Oh you mean plugging it into the calculator, or like physically solving it? \[\frac{3}{2\sqrt{1+3(1)}} \implies \frac{3}{2(1+3(1))^{1/2}}\]

  64. anonymous
    • one year ago
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    physically solving it

  65. Jhannybean
    • one year ago
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    So then you think... what does \(\sqrt{4} = ~? \)

  66. anonymous
    • one year ago
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    oh

  67. anonymous
    • one year ago
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    Now I remember :D

  68. Jhannybean
    • one year ago
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    : )

  69. Jhannybean
    • one year ago
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    Great! So you can think of it either way. See... if you had something like \(\large \sqrt[5]{x^4}\) you could rewrite this as \(\large x^{4/5}\) Do you see what I mean?

  70. anonymous
    • one year ago
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    Yes

  71. Jhannybean
    • one year ago
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    Cool \(\checkmark\). So what did you get as your limit after pluggign it all in?

  72. anonymous
    • one year ago
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    let me see

  73. anonymous
    • one year ago
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    3/4

  74. Jhannybean
    • one year ago
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    Awesome! that's correct.

  75. Jhannybean
    • one year ago
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    We got 3/5 done, but I got to call it a night. I've got some homework i need to finish up here, so sorry Ican't help youfinish the other 2 :\

  76. anonymous
    • one year ago
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    No problem,thank you for the help,much appreciated :)

  77. Jhannybean
    • one year ago
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    No problem! Good luck :)

  78. anonymous
    • one year ago
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    You too,take care :)

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