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alexthomas

  • 2 years ago

Find the limit if it exists...

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  1. alexthomas
    • 2 years ago
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    |dw:1372210588807:dw|

  2. alexthomas
    • 2 years ago
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    the radical is over the (+x) as well if not clear

  3. Jhannybean
    • 2 years ago
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    You want to multiply by the conjugate

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

  5. alexthomas
    • 2 years ago
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    i'm getting 25+x-25x^2 /radical 25x^2+x +5x

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

  7. alexthomas
    • 2 years ago
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    exactly

  8. Jhannybean
    • 2 years ago
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    mmhmm.

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

  10. alexthomas
    • 2 years ago
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    good i got that too it cant be simplified any further right?

  11. dumbcow
    • 2 years ago
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    i would use l'hopitals rule now

  12. Jhannybean
    • 2 years ago
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    Ah. Okay.

  13. Jhannybean
    • 2 years ago
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    LH Rule: \[\large \frac{ f'(x)}{g'(x)}\]

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

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

  16. alexthomas
    • 2 years ago
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    |dw:1372212305372:dw|

  17. Jhannybean
    • 2 years ago
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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 : \(\large \frac{50x+1}{2\sqrt{25x^2+x}} +5\)

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

  19. alexthomas
    • 2 years ago
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    i got the answer to be 1/10 thank you

  20. Jhannybean
    • 2 years ago
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    Can you explain to me how you got that? :o

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

  22. Jhannybean
    • 2 years ago
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    An unforseen method of approach!

  23. dan815
    • 2 years ago
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    |dw:1372214578635:dw|

  24. dan815
    • 2 years ago
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    what shud we subtract there

  25. Jhannybean
    • 2 years ago
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    Uhhh...

  26. dan815
    • 2 years ago
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    i dunno for some reason i feel like it cant be 0

  27. dan815
    • 2 years ago
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    |dw:1372215149908:dw|

  28. Jhannybean
    • 2 years ago
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    Hmmm...

  29. dan815
    • 2 years ago
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    1/10 is probably right

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

  31. dan815
    • 2 years ago
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    its tough sqrt roots arent friendly

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

  33. dan815
    • 2 years ago
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    gotta use other ways

  34. dan815
    • 2 years ago
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    i didnt im just saying that answer looks right lol

  35. Jhannybean
    • 2 years ago
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    secret methods....hidden treasures.

  36. Jhannybean
    • 2 years ago
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    I want toknow this.

  37. dan815
    • 2 years ago
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    maybe we can rewrite this as some other known function base like hyperbola?

  38. RaphaelFilgueiras
    • 2 years ago
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    |dw:1372215364565:dw|

  39. dan815
    • 2 years ago
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    oh ok

  40. Jhannybean
    • 2 years ago
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    what??...

  41. RaphaelFilgueiras
    • 2 years ago
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    dont need to use L'H,just put x in evidence

  42. Jhannybean
    • 2 years ago
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    im so confused.....wait a minute...

  43. dan815
    • 2 years ago
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    shudda known it was something simple

  44. dan815
    • 2 years ago
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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

  45. Jhannybean
    • 2 years ago
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    Oh...hm. Let me try this.

  46. RaphaelFilgueiras
    • 2 years ago
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    |dw:1372215619036:dw|

  47. dan815
    • 2 years ago
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    |dw:1372215830146:dw|

  48. dan815
    • 2 years ago
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    @whpalmer4 show us the way

  49. RaphaelFilgueiras
    • 2 years ago
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    |dw:1372215784059:dw|

  50. Jhannybean
    • 2 years ago
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    Am i wrong?

  51. RaphaelFilgueiras
    • 2 years ago
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    i think that is not right

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

  53. Jhannybean
    • 2 years ago
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    Oh YEAH. 5x/x * my bad.

  54. Jhannybean
    • 2 years ago
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    I forgot the x...... 1/10 is the answer,lol. whoops.

  55. RaphaelFilgueiras
    • 2 years ago
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    @dan815 x²(25+1/x)= 25x²+x

  56. dan815
    • 2 years ago
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    |dw:1372216059586:dw|

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

  58. RaphaelFilgueiras
    • 2 years ago
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    |dw:1372216149948:dw|

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

  60. Jhannybean
    • 2 years ago
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    When lim tends to infinity, divide by highest exponent. When lim tends to a number, use LH rule :D

  61. RaphaelFilgueiras
    • 2 years ago
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    just take care because if x goes to - inf |x|=-x

  62. Jhannybean
    • 2 years ago
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    yeah. I understand :)

  63. RaphaelFilgueiras
    • 2 years ago
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    sqrt(x²)=|x| not x; we do that only because we know x is positive(inf)

  64. Jhannybean
    • 2 years ago
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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)

  65. Jhannybean
    • 2 years ago
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    lol we're saying the same thing....2 different ways :(

  66. RaphaelFilgueiras
    • 2 years ago
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    if x goes to - inf the result isn't the same

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

  68. whpalmer4
    • 2 years ago
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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 \[=\frac{1}{5+\sqrt{25}}\]

  69. Jhannybean
    • 2 years ago
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    Oh, I see.

  70. Jhannybean
    • 2 years ago
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    Tried @whpalmer4 's way myself just now. REALLY interesting hahaha

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