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mathcalculus

  • one year ago

HELP: Evaluate the limit: lim ((1/t)-(1/8))/ (t-8) = t->8

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  1. alexwee123
    • one year ago
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    is this the problem? |dw:1374780769038:dw|

  2. kropot72
    • one year ago
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    To find the limit consider values of t less than 8 (t approaches 8 from below) and values of t greater than 8 (t approaches 8 from above) and construct a table for the value of f(t) using the following values of t: 7.9, 7.95, 7.99, 8, 8.01, 8.05, 8.1 Although f(8) is undefined, you can still find a value for the limit.

  3. mathcalculus
    • one year ago
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    can you show me?

  4. mathcalculus
    • one year ago
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    i've tried every way but i've gotten it incorrect a couple of times..

  5. mathcalculus
    • one year ago
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    @kropot72

  6. kropot72
    • one year ago
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    To start your table by finding the value of f(7.9) just plug 7.9 into the expression for f(t) \[f(7.9)=\frac{\frac{1}{7.9}-\frac{1}{8}}{7.9-8}=you\ can\ calculate\] Then find f(7.95), f(7.99), f(8.01), f(8.05) and f(8.1). The limit will become clear when you have finished your table.

  7. mathcalculus
    • one year ago
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    is there another way to do this?

  8. mathcalculus
    • one year ago
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    besides approaching it to the closest number?

  9. mathcalculus
    • one year ago
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    not the table method.

  10. mathcalculus
    • one year ago
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    @kropot72

  11. kropot72
    • one year ago
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    There is a quick method. \[\frac{\frac{1}{t}-\frac{1}{8}}{t-8}=\frac{8-t}{8t}\times\frac{-1}{8-t}=\frac{-1}{8t}\] When you plug the value t=8 into this expression you get a value for the limit. This method is not as reliable as the table method, the reason being that it does not check that the limit is aproached to the same value from above and below the limit.

  12. mathcalculus
    • one year ago
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    @kropot why did you multiply it by -1?

  13. mathcalculus
    • one year ago
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    @dumbcow could u be able to help me with this problem?

  14. mathcalculus
    • one year ago
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    @Loser66 i got 8-t/8t but then after i'm not sure how they got -1/8t

  15. mathcalculus
    • one year ago
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    you were typing? lol

  16. Loser66
    • one year ago
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    kropot72 gave you the right logic , just plug t =8 to get -1/64, done

  17. mathcalculus
    • one year ago
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    yeah i know, but i understand up to here: 8-(8)/8t... but then if i plug in 8 into this.... then i get 0/64 which is not right..... because im suppose to get a -1 on top/8(t)

  18. Loser66
    • one year ago
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    re read the stuff, you missed something, it 's not that.

  19. dumbcow
    • one year ago
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    another option is l'hopitals rule since you get 0/0 when evaluating limit differentiate top/bottom separately then reevaluate you should get lim = -1/64

  20. mathcalculus
    • one year ago
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    i know, my question is why did he multiply -1/8-t ?

  21. Loser66
    • one year ago
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    ok, dumbcow 's solution is ok, too. but kropot 72 's solution is ok, too. Yours is wrong, hehe

  22. asnaseer
    • one year ago
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    |dw:1375031598300:dw|

  23. asnaseer
    • one year ago
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    does that clarify it?

  24. Loser66
    • one year ago
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    yup, bravo @asnaseer

  25. mathcalculus
    • one year ago
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    @asnaseer yes:) just need to know why we multiply -1/8-t?

  26. mathcalculus
    • one year ago
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    @Loser66 not helping. bye

  27. Loser66
    • one year ago
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    ok, just because you tag me, I leave now, thanks for sending me away

  28. asnaseer
    • one year ago
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    @kropot72 made use of the fact that:\[\frac{1}{t-8}=\frac{-1}{8-t}\]

  29. dumbcow
    • one year ago
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    @mathcalculus , its because the denominator was "t-8" the rule when dividing is multiply by reciprocal ... which turns it into 1/t-8 then they factored out a neg

  30. mathcalculus
    • one year ago
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    ooooh okay. thank you so much @dumbcow

  31. mathcalculus
    • one year ago
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    @asnaseer thank you!

  32. asnaseer
    • one year ago
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    yw :)

  33. mathcalculus
    • one year ago
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    hold oni just multiplied 8-t/8t by 1/t-8 and i got this: 8-t/8t^2-64 @asnaseer

  34. mathcalculus
    • one year ago
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    im not sure what to do after:?

  35. asnaseer
    • one year ago
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    |dw:1375032333583:dw|

  36. asnaseer
    • one year ago
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    the terms with (8-t) cancel out

  37. asnaseer
    • one year ago
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    remember:\[t-8=-(8-t)\]therefore:\[\frac{1}{t-8}=\frac{1}{-(8-t)}=\frac{-1}{8-t}\]

  38. mathcalculus
    • one year ago
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    yeah but i can't see that when im doing this problem alone.. im thinking i multiply across.. is that possible?

  39. mathcalculus
    • one year ago
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    or cross multiply*

  40. asnaseer
    • one year ago
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    in order to evaluate the limit you need to either simplify the fraction (which is what we did), or use l'Hopitals rule (as suggested by dumbcow)

  41. mathcalculus
    • one year ago
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    but can we or no?

  42. mathcalculus
    • one year ago
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    i have not used the l hospitals rule yet.

  43. asnaseer
    • one year ago
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    if you use the first method, then you need to cancel out the common factors in the fraction, i.e. cancel out the (8-t) from the numerator and the denominator

  44. mathcalculus
    • one year ago
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    yes i see that

  45. asnaseer
    • one year ago
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    if you don't cancel these out and just multiply across then you will end up with a fraction that goes to zero divided by zero as t approaches 8

  46. mathcalculus
    • one year ago
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    i know, but all i see is that i can cross multiply but how do you know that the t-8 needed to be factored?

  47. asnaseer
    • one year ago
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    it was just a /hunch/ - it was a good candidate to try and cancel out as it tends to zero as t approaches 8.

  48. asnaseer
    • one year ago
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    and we want to avoid zeros (especially in the denominators) when evaluating limits

  49. asnaseer
    • one year ago
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    e.g. find the limit as t tends to 2 of:\[\frac{t^2-2t}{2t-4}\]

  50. asnaseer
    • one year ago
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    if you just stick t=2 into this, you will get 0/0 which is undefined

  51. asnaseer
    • one year ago
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    so first we try to factorise it and see if we get anything interesting

  52. asnaseer
    • one year ago
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    \[\frac{t^2-2t}{2t-4}=\frac{t(t-2)}{2(t-2)}\]

  53. asnaseer
    • one year ago
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    now we notice a (t-2) can be cancelled out

  54. asnaseer
    • one year ago
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    \[\frac{t^2-2t}{2t-4}=\frac{t\cancel{(t-2)}}{2\cancel{(t-2)}}=\frac{t}{2}\]

  55. asnaseer
    • one year ago
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    now we can evaluate the limit as t->2

  56. asnaseer
    • one year ago
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    hope that makes sense?

  57. mathcalculus
    • one year ago
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    okay so i did this now: 8-t/8t * 1/- (8-t)

  58. mathcalculus
    • one year ago
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    i cancel the 8-t... but what about the negative sign outside of it???

  59. mathcalculus
    • one year ago
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    im left with 1/8t

  60. asnaseer
    • one year ago
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    the negative sign can just be moved up to the 1

  61. mathcalculus
    • one year ago
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    but not -1??

  62. mathcalculus
    • one year ago
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    why??

  63. asnaseer
    • one year ago
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    |dw:1375033388098:dw|

  64. asnaseer
    • one year ago
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    all 3 forms are equivalent

  65. mathcalculus
    • one year ago
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    i need rules, not simply "oh because we can." "just guessing" "that works"

  66. mathcalculus
    • one year ago
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    oay i understand that

  67. asnaseer
    • one year ago
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    the rules here are to look for common factors that can be cancelled out

  68. mathcalculus
    • one year ago
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    so when i asked u if we can multiply across, can we or not?

  69. asnaseer
    • one year ago
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    in later classes you will be given limits where the fractions cannot be simplified. that is where you may be introduced to l'hopitals method.

  70. asnaseer
    • one year ago
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    you CAN multiply across but it won't simplify the fraction. why wouldn't you want to cancel out the common terms BEFORE multiplying?

  71. asnaseer
    • one year ago
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    e.g.: |dw:1375033624859:dw|

  72. asnaseer
    • one year ago
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    or: |dw:1375033674334:dw|

  73. mathcalculus
    • one year ago
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    okay from here:8-t/8t^2-64t

  74. asnaseer
    • one year ago
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    lead to the same simplification

  75. mathcalculus
    • one year ago
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    so how do i do this after?

  76. asnaseer
    • one year ago
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    yes - then you will have to re-factorise the denominator to cancel out the (8-t)

  77. asnaseer
    • one year ago
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    |dw:1375033741276:dw|

  78. mathcalculus
    • one year ago
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    okay i understand.

  79. asnaseer
    • one year ago
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    great! :)

  80. mathcalculus
    • one year ago
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    now when i cancel... and the negative sign is all alone. can i replace it by a negative 1 since -1/1 or 1/-1 is the same??

  81. asnaseer
    • one year ago
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    yes :)

  82. mathcalculus
    • one year ago
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    thanks! =]

  83. asnaseer
    • one year ago
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    yw :)

  84. mathcalculus
    • one year ago
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    can we go over another problem so i can feel more confident on it? @asneer

  85. asnaseer
    • one year ago
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    sure - just post it as a new question please

  86. mathcalculus
    • one year ago
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  87. mathcalculus
    • one year ago
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    k

  88. asnaseer
    • one year ago
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    <<<--- please post as a new question in the list on the left it helps others learn as well :)

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