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anonymous

  • 4 years ago

find the limit of 3x^2+6x-9/2x^2-2 as x approaches 1

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  1. asnaseer
    • 4 years ago
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    You might find this site useful: http://archives.math.utk.edu/visual.calculus/1/limits.7/ It has good walk-through tutorials on this topic.

  2. anonymous
    • 4 years ago
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    you basically just need to brush up on factoring

  3. anonymous
    • 4 years ago
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    (a+b)^2=a^2+2ab+b^2, if you factor a 3 out of that numerator, you will have this situation

  4. anonymous
    • 4 years ago
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    yeah how do i factor 9 to add up to 6

  5. anonymous
    • 4 years ago
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    or do i not have to

  6. anonymous
    • 4 years ago
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    factor out a 3 first, it will make it much easier

  7. anonymous
    • 4 years ago
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    ive got (x+3)(x-3)

  8. anonymous
    • 4 years ago
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    thats wrong

  9. anonymous
    • 4 years ago
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    thats a difference of 2 squares

  10. anonymous
    • 4 years ago
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    so what do i do

  11. anonymous
    • 4 years ago
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    if you expanded that you'd get x^2-9

  12. anonymous
    • 4 years ago
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    this part 3x^2+6x-9

  13. anonymous
    • 4 years ago
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    factor out a 3

  14. anonymous
    • 4 years ago
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    3,6,9 all have a common factor of 3 here

  15. anonymous
    • 4 years ago
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    3(x^2 +6x-9)

  16. anonymous
    • 4 years ago
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    you have to factor the 3 out of the whole numerator so that when you multiply it you get the original expression back

  17. anonymous
    • 4 years ago
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    what you just did will yield 3x^2 + 18x -27

  18. anonymous
    • 4 years ago
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    3(x^2-2x-3)

  19. anonymous
    • 4 years ago
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    +2x

  20. anonymous
    • 4 years ago
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    good

  21. anonymous
    • 4 years ago
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    now factor the denominator

  22. anonymous
    • 4 years ago
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    very much the same thing, guess which number you're pulling out?

  23. anonymous
    • 4 years ago
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    2(x^2-1)

  24. anonymous
    • 4 years ago
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    yes

  25. anonymous
    • 4 years ago
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    now factor (x^2-1)

  26. anonymous
    • 4 years ago
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    (difference of 2 squares), i mentioned it a little bit ago

  27. anonymous
    • 4 years ago
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    hmm im not sure here

  28. anonymous
    • 4 years ago
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    hehe you just did it mistakenly earlier

  29. anonymous
    • 4 years ago
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    remember you typed (x-3)(x+3)

  30. anonymous
    • 4 years ago
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    oh that

  31. anonymous
    • 4 years ago
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    (x+1)(x-1)

  32. anonymous
    • 4 years ago
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    yes, now it's all factored. put it all together and cancel like terms

  33. anonymous
    • 4 years ago
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    ok but could i have plug in 1 and get an answer

  34. anonymous
    • 4 years ago
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    actually, you can further factor the numerator

  35. anonymous
    • 4 years ago
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    right now you have the numerator at (x^2+2x-3)

  36. anonymous
    • 4 years ago
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    you could not have p lugged 1 into the original equation before factoring it

  37. anonymous
    • 4 years ago
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    look at the original denominator

  38. anonymous
    • 4 years ago
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    2x^2-2

  39. anonymous
    • 4 years ago
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    if you plugged a 1 in here, you'd get a divide by 0 error

  40. anonymous
    • 4 years ago
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    i see you will just get 0/0

  41. anonymous
    • 4 years ago
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    so try factoring x^2+2x-3

  42. anonymous
    • 4 years ago
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    (x-1)(x+3)

  43. anonymous
    • 4 years ago
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    good

  44. anonymous
    • 4 years ago
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    now everything is factored

  45. anonymous
    • 4 years ago
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    so when i cancel i get x+3/x+1

  46. anonymous
    • 4 years ago
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    don't forget about the 3 and 2 you factored out

  47. anonymous
    • 4 years ago
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    3(x+3)/(2(x+1)

  48. anonymous
    • 4 years ago
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    at this point you can plug in 1 and you will not get a divide by 0 error

  49. anonymous
    • 4 years ago
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    because you cancelled out the (x-1)

  50. anonymous
    • 4 years ago
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    ok i get 12/4

  51. anonymous
    • 4 years ago
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    when i reduce i get 3/1

  52. anonymous
    • 4 years ago
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    yes. that's it

  53. anonymous
    • 4 years ago
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    thanks for the help

  54. anonymous
    • 4 years ago
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    3

  55. anonymous
    • 4 years ago
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    np

  56. anonymous
    • 4 years ago
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    i have another limit problem but its approaching infinity

  57. anonymous
    • 4 years ago
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    ok, let's have it

  58. anonymous
    • 4 years ago
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    x^2 + 2x-3/2x^2-2 as x approaches infinity, what is the limit

  59. anonymous
    • 4 years ago
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    now look at this numerator and denominator

  60. anonymous
    • 4 years ago
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    you already factored them in the last problem

  61. anonymous
    • 4 years ago
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    give me your factored equation for this problem

  62. anonymous
    • 4 years ago
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    (x-1)(x+3)/2(x^2-1)

  63. anonymous
    • 4 years ago
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    the denominator further factors out to (x-1)(x+1)

  64. anonymous
    • 4 years ago
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    yes, now cancel

  65. anonymous
    • 4 years ago
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    x+1/x+1

  66. anonymous
    • 4 years ago
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    or 2x+1

  67. anonymous
    • 4 years ago
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    what happened to x+3?

  68. anonymous
    • 4 years ago
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    my final looks like this (x+3)/(2x+1)

  69. anonymous
    • 4 years ago
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    yes

  70. anonymous
    • 4 years ago
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    2x+2

  71. anonymous
    • 4 years ago
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    no

  72. anonymous
    • 4 years ago
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    2(x+1) then you foil

  73. anonymous
    • 4 years ago
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    that's as far as you can factor it: (x+3)/[2(x+1)]

  74. anonymous
    • 4 years ago
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    you cant do 2x+2?

  75. anonymous
    • 4 years ago
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    what's your thinking on that? just trying to see where you're going wrong

  76. anonymous
    • 4 years ago
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    oh my mistake, i see what you were doing

  77. anonymous
    • 4 years ago
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    thats not a big deal to expand the denominator to 2x+2

  78. anonymous
    • 4 years ago
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    so youd have (x+3)/(2x+2)

  79. anonymous
    • 4 years ago
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    i thought you meant the whole equation factored out to that

  80. anonymous
    • 4 years ago
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    if you have 2(x+1)...why not just foil (2)(x)=2x...2(1)=2...2x+2...

  81. anonymous
    • 4 years ago
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    you can, but it's generally not necessary after you've already factored

  82. anonymous
    • 4 years ago
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    plus if you do it too early you'll have to factor it again later

  83. anonymous
    • 4 years ago
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    oh ok

  84. anonymous
    • 4 years ago
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    so since i can't plug in infinity...what should i do

  85. anonymous
    • 4 years ago
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    now you have to think what would happen to f(x) is you kept plugging in very large numbers

  86. anonymous
    • 4 years ago
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    it would remain large

  87. anonymous
    • 4 years ago
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    nope

  88. anonymous
    • 4 years ago
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    try plugging in 100 just to test it

  89. anonymous
    • 4 years ago
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    well it might cut down...but at the end it would still be large

  90. anonymous
    • 4 years ago
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    100+3/2(100+1)=103/202= .50

  91. anonymous
    • 4 years ago
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    ok now 1000

  92. anonymous
    • 4 years ago
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    the same .50

  93. anonymous
    • 4 years ago
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    ok

  94. anonymous
    • 4 years ago
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    so 1/2?

  95. anonymous
    • 4 years ago
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    1 sec

  96. anonymous
    • 4 years ago
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    that's right

  97. anonymous
    • 4 years ago
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    ok thanks for the help

  98. anonymous
    • 4 years ago
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    sure

  99. anonymous
    • 4 years ago
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    just understand why that's the answer

  100. anonymous
    • 4 years ago
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    when you start using very large numbers, +3 in the numerator and the +2 in the denominator are pretty much meaningless

  101. anonymous
    • 4 years ago
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    essentially its like graphing x/(2x)

  102. anonymous
    • 4 years ago
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    ok

  103. anonymous
    • 4 years ago
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    thanks for the explanaition

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