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mathcalculus Group Title

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

  • one year ago
  • one year ago

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  1. alexwee123 Group Title
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    is this the problem? |dw:1374780769038:dw|

    • one year ago
  2. kropot72 Group Title
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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.

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

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

    • one year ago
  5. mathcalculus Group Title
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    @kropot72

    • one year ago
  6. kropot72 Group Title
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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.

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

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

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

    • one year ago
  10. mathcalculus Group Title
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    @kropot72

    • one year ago
  11. kropot72 Group Title
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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.

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

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

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

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

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

    • one year ago
  17. mathcalculus Group Title
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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)

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

    • one year ago
  19. dumbcow Group Title
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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

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

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

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

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

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

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

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

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

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

    • one year ago
  29. dumbcow Group Title
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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

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

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

    • one year ago
  32. asnaseer Group Title
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    yw :)

    • one year ago
  33. mathcalculus Group Title
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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

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

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

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

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

    • one year ago
  38. mathcalculus Group Title
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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?

    • one year ago
  39. mathcalculus Group Title
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    or cross multiply*

    • one year ago
  40. asnaseer Group Title
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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)

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

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

    • one year ago
  43. asnaseer Group Title
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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

    • one year ago
  44. mathcalculus Group Title
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    yes i see that

    • one year ago
  45. asnaseer Group Title
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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

    • one year ago
  46. mathcalculus Group Title
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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?

    • one year ago
  47. asnaseer Group Title
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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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    • one year ago
  62. mathcalculus Group Title
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    why??

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

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

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

    • one year ago
  66. mathcalculus Group Title
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    oay i understand that

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

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

    • one year ago
  69. asnaseer Group Title
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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.

    • one year ago
  70. asnaseer Group Title
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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?

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

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

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

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

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

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

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

    • one year ago
  78. mathcalculus Group Title
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    okay i understand.

    • one year ago
  79. asnaseer Group Title
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    great! :)

    • one year ago
  80. mathcalculus Group Title
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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??

    • one year ago
  81. asnaseer Group Title
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    yes :)

    • one year ago
  82. mathcalculus Group Title
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    thanks! =]

    • one year ago
  83. asnaseer Group Title
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    yw :)

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

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

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

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

    • one year ago
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