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alexthomas Group Title

Find the limit if it exists...

  • one year ago
  • one year ago

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  1. alexthomas Group Title
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    |dw:1372210588807:dw|

    • one year ago
  2. alexthomas Group Title
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    the radical is over the (+x) as well if not clear

    • one year ago
  3. Jhannybean Group Title
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    You want to multiply by the conjugate

    • one year ago
  4. Jhannybean Group Title
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    \[\large \lim_{x \rightarrow \infty} \ \frac{\sqrt{25x^2+x}-5x}{1} \cdot \frac{\sqrt{25x^2+x} +5x}{\sqrt{25x^2+x} +5x}\]

    • one year ago
  5. alexthomas Group Title
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    i'm getting 25+x-25x^2 /radical 25x^2+x +5x

    • one year ago
  6. Jhannybean Group Title
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    \[\large \lim_{x \rightarrow \infty} \ \frac{(25x^2+x)-(5x)^2}{\sqrt{25x^2+x} +5x} \]

    • one year ago
  7. alexthomas Group Title
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    exactly

    • one year ago
  8. Jhannybean Group Title
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    mmhmm.

    • one year ago
  9. Jhannybean Group Title
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    \[\large \lim_{x \rightarrow \infty} \ \frac{x}{\sqrt{25x^2+x} +5x}\]

    • one year ago
  10. alexthomas Group Title
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    good i got that too it cant be simplified any further right?

    • one year ago
  11. dumbcow Group Title
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    i would use l'hopitals rule now

    • one year ago
  12. Jhannybean Group Title
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    Ah. Okay.

    • one year ago
  13. Jhannybean Group Title
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    LH Rule: \[\large \frac{ f'(x)}{g'(x)}\]

    • one year ago
  14. Jhannybean Group Title
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    Since if we took the limit to infinity we'd get \(\large \frac{\infty}{\infty}\)

    • one year ago
  15. Jhannybean Group Title
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    Can you tell me what the derivative of \(\large \sqrt{25x^2 +x} +5x\) is?

    • one year ago
  16. alexthomas Group Title
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    |dw:1372212305372:dw|

    • one year ago
  17. Jhannybean Group Title
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    Not exactly.. We have \(\large \sqrt{25x^2+x}+5x\) right? That can be rewritten as \(\large (25x^2 +x)^{1/2}+5x\) Using chain rule: \(\large \frac{1}{2\sqrt{25x^2+x}} \cdot (50x +1) +5\) Simplify : \(\large \frac{50x+1}{2\sqrt{25x^2+x}} +5\)

    • one year ago
  18. Jhannybean Group Title
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    So now we'll have... \[\huge \lim_{x \rightarrow \infty} \ \frac{\frac{1}{50x+1}}{2\sqrt{25x^2+x}+5}\]

    • one year ago
  19. alexthomas Group Title
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    i got the answer to be 1/10 thank you

    • one year ago
  20. Jhannybean Group Title
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    Can you explain to me how you got that? :o

    • one year ago
  21. Jhannybean Group Title
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    what i got was \[\large \lim_{x \rightarrow \infty} \ \frac{1}{(100x+2)(\sqrt{25x^2+x}+5)} = 0 \]

    • one year ago
  22. Jhannybean Group Title
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    An unforseen method of approach!

    • one year ago
  23. dan815 Group Title
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    |dw:1372214578635:dw|

    • one year ago
  24. dan815 Group Title
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    what shud we subtract there

    • one year ago
  25. Jhannybean Group Title
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    Uhhh...

    • one year ago
  26. dan815 Group Title
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    i dunno for some reason i feel like it cant be 0

    • one year ago
  27. dan815 Group Title
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    |dw:1372215149908:dw|

    • one year ago
  28. Jhannybean Group Title
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    Hmmm...

    • one year ago
  29. dan815 Group Title
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    1/10 is probably right

    • one year ago
  30. Jhannybean Group Title
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    How do you factor out a -5x from \(\large \sqrt{25x^2+x}\) ?

    • one year ago
  31. dan815 Group Title
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    its tough sqrt roots arent friendly

    • one year ago
  32. Jhannybean Group Title
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    HOW DO YOU GET 1/10???? thats what i wanted to know! and then the guy disappeared lol.

    • one year ago
  33. dan815 Group Title
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    gotta use other ways

    • one year ago
  34. dan815 Group Title
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    i didnt im just saying that answer looks right lol

    • one year ago
  35. Jhannybean Group Title
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    secret methods....hidden treasures.

    • one year ago
  36. Jhannybean Group Title
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    I want toknow this.

    • one year ago
  37. dan815 Group Title
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    maybe we can rewrite this as some other known function base like hyperbola?

    • one year ago
  38. RaphaelFilgueiras Group Title
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    |dw:1372215364565:dw|

    • one year ago
  39. dan815 Group Title
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    oh ok

    • one year ago
  40. Jhannybean Group Title
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    what??...

    • one year ago
  41. RaphaelFilgueiras Group Title
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    dont need to use L'H,just put x in evidence

    • one year ago
  42. Jhannybean Group Title
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    im so confused.....wait a minute...

    • one year ago
  43. dan815 Group Title
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    shudda known it was something simple

    • one year ago
  44. dan815 Group Title
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    basically hes saying the highest exponents there will just be root(25x^2) and 5x so we can still divide everything by x

    • one year ago
  45. Jhannybean Group Title
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    Oh...hm. Let me try this.

    • one year ago
  46. RaphaelFilgueiras Group Title
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    |dw:1372215619036:dw|

    • one year ago
  47. dan815 Group Title
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    |dw:1372215830146:dw|

    • one year ago
  48. dan815 Group Title
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    @whpalmer4 show us the way

    • one year ago
  49. RaphaelFilgueiras Group Title
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    |dw:1372215784059:dw|

    • one year ago
  50. Jhannybean Group Title
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    Am i wrong?

    • one year ago
  51. RaphaelFilgueiras Group Title
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    i think that is not right

    • one year ago
  52. dan815 Group Title
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    u shud have +5 there instead of last 0, and ur diving by x that is the higest power

    • one year ago
  53. Jhannybean Group Title
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    Oh YEAH. 5x/x * my bad.

    • one year ago
  54. Jhannybean Group Title
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    I forgot the x...... 1/10 is the answer,lol. whoops.

    • one year ago
  55. RaphaelFilgueiras Group Title
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    @dan815 x²(25+1/x)= 25x²+x

    • one year ago
  56. dan815 Group Title
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    |dw:1372216059586:dw|

    • one year ago
  57. whpalmer4 Group Title
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    you were on the right track \[\large \lim_{x \rightarrow \infty} \ \frac{x}{\sqrt{25x^2+x} +5x} = \large \lim_{x \rightarrow \infty} \ \frac{1}{\frac{\sqrt{25x^2+x} +5x}{x}}\]

    • one year ago
  58. RaphaelFilgueiras Group Title
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    |dw:1372216149948:dw|

    • one year ago
  59. Jhannybean Group Title
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    \[\large \lim_{x \rightarrow \infty} \ \frac{x/x}{\sqrt{25x^2/x^2 +x/x^2}+5x/x}\]\[\large \frac{1}{\sqrt{25+0}+5} =\frac{1}{10} \]

    • one year ago
  60. Jhannybean Group Title
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    When lim tends to infinity, divide by highest exponent. When lim tends to a number, use LH rule :D

    • one year ago
  61. RaphaelFilgueiras Group Title
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    just take care because if x goes to - inf |x|=-x

    • one year ago
  62. Jhannybean Group Title
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    yeah. I understand :)

    • one year ago
  63. RaphaelFilgueiras Group Title
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    sqrt(x²)=|x| not x; we do that only because we know x is positive(inf)

    • one year ago
  64. Jhannybean Group Title
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    You divide everything under a square root by the highest power underneath the root, and all the other x's get divided by the value of x that is dependent on the behavior of the function, (i.e if x tends to positive or negative infinity)

    • one year ago
  65. Jhannybean Group Title
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    lol we're saying the same thing....2 different ways :(

    • one year ago
  66. RaphaelFilgueiras Group Title
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    if x goes to - inf the result isn't the same

    • one year ago
  67. Jhannybean Group Title
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    yes,so if x goes to -inf, x takes on neg values.... hence we divide by -x instead of x.

    • one year ago
  68. whpalmer4 Group Title
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    \[\large \large \lim_{x \rightarrow \infty} \ \frac{1}{\frac{\sqrt{25x^2+x} +5x}{x}} = \frac{1}{5+ \sqrt{\lim_{x\rightarrow \infty} \ \frac{25x^2+x}{x^2}}} \]\[= \frac{1}{5+ \sqrt{\lim_{x\rightarrow \infty} \ \frac{50x+1}{2x}}}\]by L'Hopital \[=\frac{1}{5+\sqrt{25}}\]

    • one year ago
  69. Jhannybean Group Title
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    Oh, I see.

    • one year ago
  70. Jhannybean Group Title
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    Tried @whpalmer4 's way myself just now. REALLY interesting hahaha

    • one year ago
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