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mathcalculus

  • 2 years ago

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

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  1. alexwee123
    • 2 years ago
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    is this the problem? |dw:1374780769038:dw|

  2. kropot72
    • 2 years 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
    • 2 years ago
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    can you show me?

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

  5. mathcalculus
    • 2 years ago
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    @kropot72

  6. kropot72
    • 2 years 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
    • 2 years ago
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    is there another way to do this?

  8. mathcalculus
    • 2 years ago
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    besides approaching it to the closest number?

  9. mathcalculus
    • 2 years ago
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    not the table method.

  10. mathcalculus
    • 2 years ago
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    @kropot72

  11. kropot72
    • 2 years 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
    • 2 years ago
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    @kropot why did you multiply it by -1?

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

  14. mathcalculus
    • 2 years 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
    • 2 years ago
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    you were typing? lol

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

  17. mathcalculus
    • 2 years 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
    • 2 years ago
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    re read the stuff, you missed something, it 's not that.

  19. dumbcow
    • 2 years 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
    • 2 years ago
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    i know, my question is why did he multiply -1/8-t ?

  21. Loser66
    • 2 years 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
    • 2 years ago
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    |dw:1375031598300:dw|

  23. asnaseer
    • 2 years ago
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    does that clarify it?

  24. Loser66
    • 2 years ago
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    yup, bravo @asnaseer

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

  26. mathcalculus
    • 2 years ago
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    @Loser66 not helping. bye

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

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

  29. dumbcow
    • 2 years 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
    • 2 years ago
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    ooooh okay. thank you so much @dumbcow

  31. mathcalculus
    • 2 years ago
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    @asnaseer thank you!

  32. asnaseer
    • 2 years ago
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    yw :)

  33. mathcalculus
    • 2 years 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
    • 2 years ago
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    im not sure what to do after:?

  35. asnaseer
    • 2 years ago
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    |dw:1375032333583:dw|

  36. asnaseer
    • 2 years ago
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    the terms with (8-t) cancel out

  37. asnaseer
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    or cross multiply*

  40. asnaseer
    • 2 years 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
    • 2 years ago
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    but can we or no?

  42. mathcalculus
    • 2 years ago
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    i have not used the l hospitals rule yet.

  43. asnaseer
    • 2 years 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
    • 2 years ago
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    yes i see that

  45. asnaseer
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    and we want to avoid zeros (especially in the denominators) when evaluating limits

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

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

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

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

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

  54. asnaseer
    • 2 years 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
    • 2 years ago
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    now we can evaluate the limit as t->2

  56. asnaseer
    • 2 years ago
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    hope that makes sense?

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

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

  59. mathcalculus
    • 2 years ago
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    im left with 1/8t

  60. asnaseer
    • 2 years ago
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    the negative sign can just be moved up to the 1

  61. mathcalculus
    • 2 years ago
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    but not -1??

  62. mathcalculus
    • 2 years ago
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    why??

  63. asnaseer
    • 2 years ago
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    |dw:1375033388098:dw|

  64. asnaseer
    • 2 years ago
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    all 3 forms are equivalent

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

  66. mathcalculus
    • 2 years ago
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    oay i understand that

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

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

  69. asnaseer
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    e.g.: |dw:1375033624859:dw|

  72. asnaseer
    • 2 years ago
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    or: |dw:1375033674334:dw|

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

  74. asnaseer
    • 2 years ago
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    lead to the same simplification

  75. mathcalculus
    • 2 years ago
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    so how do i do this after?

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

  77. asnaseer
    • 2 years ago
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    |dw:1375033741276:dw|

  78. mathcalculus
    • 2 years ago
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    okay i understand.

  79. asnaseer
    • 2 years ago
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    great! :)

  80. mathcalculus
    • 2 years 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
    • 2 years ago
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    yes :)

  82. mathcalculus
    • 2 years ago
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    thanks! =]

  83. asnaseer
    • 2 years ago
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    yw :)

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

  85. asnaseer
    • 2 years ago
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    sure - just post it as a new question please

  86. mathcalculus
    • 2 years ago
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  87. mathcalculus
    • 2 years ago
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    k

  88. asnaseer
    • 2 years 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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