## mathcalculus Group Title HELP: Evaluate the limit: lim ((1/t)-(1/8))/ (t-8) = t->8 one year ago one year ago

1. alexwee123 Group Title

is this the problem? |dw:1374780769038:dw|

2. kropot72 Group Title

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 Group Title

can you show me?

4. mathcalculus Group Title

i've tried every way but i've gotten it incorrect a couple of times..

5. mathcalculus Group Title

@kropot72

6. kropot72 Group Title

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 Group Title

is there another way to do this?

8. mathcalculus Group Title

besides approaching it to the closest number?

9. mathcalculus Group Title

not the table method.

10. mathcalculus Group Title

@kropot72

11. kropot72 Group Title

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 Group Title

@kropot why did you multiply it by -1?

13. mathcalculus Group Title

@dumbcow could u be able to help me with this problem?

14. mathcalculus Group Title

@Loser66 i got 8-t/8t but then after i'm not sure how they got -1/8t

15. mathcalculus Group Title

you were typing? lol

16. Loser66 Group Title

kropot72 gave you the right logic , just plug t =8 to get -1/64, done

17. mathcalculus Group Title

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 Group Title

re read the stuff, you missed something, it 's not that.

19. dumbcow Group Title

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 Group Title

i know, my question is why did he multiply -1/8-t ?

21. Loser66 Group Title

ok, dumbcow 's solution is ok, too. but kropot 72 's solution is ok, too. Yours is wrong, hehe

22. asnaseer Group Title

|dw:1375031598300:dw|

23. asnaseer Group Title

does that clarify it?

24. Loser66 Group Title

yup, bravo @asnaseer

25. mathcalculus Group Title

@asnaseer yes:) just need to know why we multiply -1/8-t?

26. mathcalculus Group Title

@Loser66 not helping. bye

27. Loser66 Group Title

ok, just because you tag me, I leave now, thanks for sending me away

28. asnaseer Group Title

@kropot72 made use of the fact that:$\frac{1}{t-8}=\frac{-1}{8-t}$

29. dumbcow Group Title

@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 Group Title

ooooh okay. thank you so much @dumbcow

31. mathcalculus Group Title

@asnaseer thank you!

32. asnaseer Group Title

yw :)

33. mathcalculus Group Title

hold oni just multiplied 8-t/8t by 1/t-8 and i got this: 8-t/8t^2-64 @asnaseer

34. mathcalculus Group Title

im not sure what to do after:?

35. asnaseer Group Title

|dw:1375032333583:dw|

36. asnaseer Group Title

the terms with (8-t) cancel out

37. asnaseer Group Title

remember:$t-8=-(8-t)$therefore:$\frac{1}{t-8}=\frac{1}{-(8-t)}=\frac{-1}{8-t}$

38. mathcalculus Group Title

yeah but i can't see that when im doing this problem alone.. im thinking i multiply across.. is that possible?

39. mathcalculus Group Title

or cross multiply*

40. asnaseer Group Title

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 Group Title

but can we or no?

42. mathcalculus Group Title

i have not used the l hospitals rule yet.

43. asnaseer Group Title

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 Group Title

yes i see that

45. asnaseer Group Title

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 Group Title

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 Group Title

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 Group Title

and we want to avoid zeros (especially in the denominators) when evaluating limits

49. asnaseer Group Title

e.g. find the limit as t tends to 2 of:$\frac{t^2-2t}{2t-4}$

50. asnaseer Group Title

if you just stick t=2 into this, you will get 0/0 which is undefined

51. asnaseer Group Title

so first we try to factorise it and see if we get anything interesting

52. asnaseer Group Title

$\frac{t^2-2t}{2t-4}=\frac{t(t-2)}{2(t-2)}$

53. asnaseer Group Title

now we notice a (t-2) can be cancelled out

54. asnaseer Group Title

$\frac{t^2-2t}{2t-4}=\frac{t\cancel{(t-2)}}{2\cancel{(t-2)}}=\frac{t}{2}$

55. asnaseer Group Title

now we can evaluate the limit as t->2

56. asnaseer Group Title

hope that makes sense?

57. mathcalculus Group Title

okay so i did this now: 8-t/8t * 1/- (8-t)

58. mathcalculus Group Title

i cancel the 8-t... but what about the negative sign outside of it???

59. mathcalculus Group Title

im left with 1/8t

60. asnaseer Group Title

the negative sign can just be moved up to the 1

61. mathcalculus Group Title

but not -1??

62. mathcalculus Group Title

why??

63. asnaseer Group Title

|dw:1375033388098:dw|

64. asnaseer Group Title

all 3 forms are equivalent

65. mathcalculus Group Title

i need rules, not simply "oh because we can." "just guessing" "that works"

66. mathcalculus Group Title

oay i understand that

67. asnaseer Group Title

the rules here are to look for common factors that can be cancelled out

68. mathcalculus Group Title

so when i asked u if we can multiply across, can we or not?

69. asnaseer Group Title

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 Group Title

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 Group Title

e.g.: |dw:1375033624859:dw|

72. asnaseer Group Title

or: |dw:1375033674334:dw|

73. mathcalculus Group Title

okay from here:8-t/8t^2-64t

74. asnaseer Group Title

75. mathcalculus Group Title

so how do i do this after?

76. asnaseer Group Title

yes - then you will have to re-factorise the denominator to cancel out the (8-t)

77. asnaseer Group Title

|dw:1375033741276:dw|

78. mathcalculus Group Title

okay i understand.

79. asnaseer Group Title

great! :)

80. mathcalculus Group Title

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 Group Title

yes :)

82. mathcalculus Group Title

thanks! =]

83. asnaseer Group Title

yw :)

84. mathcalculus Group Title

can we go over another problem so i can feel more confident on it? @asneer

85. asnaseer Group Title

sure - just post it as a new question please

86. mathcalculus Group Title

87. mathcalculus Group Title

k

88. asnaseer Group Title

<<<--- please post as a new question in the list on the left it helps others learn as well :)