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burhan101

  • 2 years ago

limit

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  1. burhan101
    • 2 years ago
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    |dw:1350454227197:dw|

  2. burhan101
    • 2 years ago
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    |dw:1350454341729:dw|

  3. terenzreignz
    • 2 years ago
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    Lol, derivative?

  4. burhan101
    • 2 years ago
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    yes i think so that doesnt cancel out tho -.- im making a mistake

  5. terenzreignz
    • 2 years ago
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    \[\huge \lim_{h \rightarrow 0}\frac{[(x+h)^{3}-3]-(x^{3}-3)}{h}\] is this it?

  6. burhan101
    • 2 years ago
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    yupp ! how do you make it big like that btw ?

  7. terenzreignz
    • 2 years ago
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    uhh, before typing in your equations and stuff, put in \huge

  8. burhan101
    • 2 years ago
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    thanks:D

  9. terenzreignz
    • 2 years ago
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    This looks like you're taking the derivative of \[\huge x^{3} - 3\]

  10. burhan101
    • 2 years ago
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    yup that's right !

  11. terenzreignz
    • 2 years ago
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    And you're to get the derivative using limits? lol, that's harsh... :D

  12. burhan101
    • 2 years ago
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    yess i have to

  13. terenzreignz
    • 2 years ago
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    You'll realise soon that there are faster ways of getting the derivative... anyway, just evaluate the numerator and show me what you get...

  14. Algebraic!
    • 2 years ago
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    x^3 's cancel 3's cancel... what's left?

  15. Algebraic!
    • 2 years ago
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    3x^2h + 3xh^2 +h^3 all divided by h factor out an 'h' in the numerator cancel with the 'h' in the denom. now you can take the limit without issues.

  16. burhan101
    • 2 years ago
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    \[\huge \ \frac{ x^3+3x^2h+3xh^2-3-x^2+3 }{h }\]

  17. Algebraic!
    • 2 years ago
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    no.

  18. burhan101
    • 2 years ago
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    i have an x^3 and x^2 ? :S

  19. Algebraic!
    • 2 years ago
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    no.

  20. terenzreignz
    • 2 years ago
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    Well that changes everything...

  21. Algebraic!
    • 2 years ago
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    \[\frac{ f(x+h) - f(x) }{h }\]

  22. Algebraic!
    • 2 years ago
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    what's f(x)?

  23. Algebraic!
    • 2 years ago
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    x^3 -3

  24. Algebraic!
    • 2 years ago
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    \[\frac{(x+h)^3 -3 - ( x^3 -3) }{h }\]

  25. burhan101
    • 2 years ago
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    no f(x)=x^2-3

  26. burhan101
    • 2 years ago
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    thats why it doesnt cancel out

  27. Algebraic!
    • 2 years ago
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    you expanded correctly: \[\frac{ x^3 +3x^2h +3xh^2 +h^3 -3 - (x^3 -3) }{h }\]

  28. Algebraic!
    • 2 years ago
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    ok

  29. Algebraic!
    • 2 years ago
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    then why are you cubing (x+h) ?

  30. burhan101
    • 2 years ago
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    ohhh i should square it *facepalm*

  31. Algebraic!
    • 2 years ago
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    \[\frac{ (x+h)^2 -3 -(x^2-3) }{h }\]

  32. burhan101
    • 2 years ago
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    can you help me with this one: find the equation of the tangent \[y=\frac{ x-2 }{ x+2 }\] when x=0

  33. burhan101
    • 2 years ago
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    \[\huge \lim_{h \rightarrow 0} \frac{ (x+h)-2 }{ (x+h)+2 } -\frac{ x-2 }{ x+2 }\]

  34. burhan101
    • 2 years ago
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    @terenzreignz

  35. terenzreignz
    • 2 years ago
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    Equation of the tangent involves derivatives, right?

  36. burhan101
    • 2 years ago
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    yes

  37. terenzreignz
    • 2 years ago
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    Quotient rule? Ever dabbled with it?

  38. burhan101
    • 2 years ago
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    no we're not allowed to use that yet -.-

  39. terenzreignz
    • 2 years ago
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    Here, I don't think I have the attention span to type this out o.O http://www.math.hmc.edu/calculus/tutorials/quotient_rule/proof.pdf

  40. terenzreignz
    • 2 years ago
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    You're allowed to use anything as long as you can prove it ;)

  41. burhan101
    • 2 years ago
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    my tests always state "Only using method thought in class" i know that quotient rule is so much faster

  42. burhan101
    • 2 years ago
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    and we we're never thought that in class yet

  43. terenzreignz
    • 2 years ago
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    Well, don't despair :D I'll see what I can do... To get the equation of the tangent line, first we need a slope. to get that slope, we need the derivative. So what's the derivative (go ahead and express it first in limit-form)

  44. burhan101
    • 2 years ago
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    \[\huge \lim_{h \rightarrow o}\frac{ (x+h)-2 }{ (x+h)+2 } - \frac{ x-2 }{ x+2 }\]

  45. terenzreignz
    • 2 years ago
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    nope... you forgot that all that must also be divided by h... small details, but no less important...

  46. burhan101
    • 2 years ago
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    ohh i did :$

  47. terenzreignz
    • 2 years ago
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    tsk \[\huge \lim_{h \rightarrow 0}\frac{\frac{ (x+h)-2 }{ (x+h)+2 } - \frac{ x-2 }{ x+2 }}{h}\]

  48. burhan101
    • 2 years ago
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    ok yupp

  49. terenzreignz
    • 2 years ago
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    Right, well this can also be written as \[\lim_{h \rightarrow 0}\frac{1}{h}\left( \frac{ (x+h)-2 }{ (x+h)+2 } - \frac{ x-2 }{ x+2 } \right)\]

  50. burhan101
    • 2 years ago
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    oh okayy

  51. terenzreignz
    • 2 years ago
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    now \[\lim_{h \rightarrow 0}\frac{1}{h}\frac{\left[ \left( x+h \right)-2 \right]\left( x+2 \right)-\left[ \left( x+h \right)+2 \right]\left( x-2 \right)}{\left[ \left( x+h \right)+2 \right]\left( x+2 \right)}\] o.O

  52. burhan101
    • 2 years ago
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    wait, how did you get rid of those fractions ?

  53. terenzreignz
    • 2 years ago
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    Like magic :P \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\]

  54. terenzreignz
    • 2 years ago
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    Get it now?

  55. burhan101
    • 2 years ago
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    yeaaah ! i do

  56. burhan101
    • 2 years ago
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    and then cancelling

  57. terenzreignz
    • 2 years ago
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    Hey, there's nothing you can cancel...

  58. burhan101
    • 2 years ago
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    then expansion and cancelling out the like terms

  59. terenzreignz
    • 2 years ago
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    Nothing that would you do you much good, anyway... Here's a tricky bit of manipulation \[\lim_{h \rightarrow 0}\frac{1}{h}\frac{\left[ \left( x+h \right)-2 \right]\left( x+2 \right)-(x-2)(x+2)+(x-2)(x+2)-\left[ \left( x+h \right)+2 \right]\left( x-2 \right)}{\left[ \left( x+h \right)+2 \right]\left( x+2 \right)}\]

  60. burhan101
    • 2 years ago
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    wow o.O

  61. terenzreignz
    • 2 years ago
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    All right, scratch everything, I overlooked one important detail...

  62. terenzreignz
    • 2 years ago
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    You only need to get its slope on the specific case where x = 0 right?

  63. terenzreignz
    • 2 years ago
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    Use this definition of derivative instead: \[f'(c)=\lim_{x \rightarrow c}\frac{f(x)-f(c)}{x-c}\]

  64. burhan101
    • 2 years ago
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    whats c ?

  65. terenzreignz
    • 2 years ago
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    Sorry for this extremely late reply, I was busy with something :( c is 0.

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