## mathcalculus 2 years ago HELP: Evaluate the limit: lim ((1/t)-(1/8))/ (t-8) = t->8

1. alexwee123

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

2. kropot72

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

can you show me?

4. mathcalculus

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

5. mathcalculus

@kropot72

6. kropot72

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

is there another way to do this?

8. mathcalculus

besides approaching it to the closest number?

9. mathcalculus

not the table method.

10. mathcalculus

@kropot72

11. kropot72

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

@kropot why did you multiply it by -1?

13. mathcalculus

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

14. mathcalculus

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

15. mathcalculus

you were typing? lol

16. Loser66

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

17. mathcalculus

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

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

19. dumbcow

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

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

21. Loser66

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

22. asnaseer

|dw:1375031598300:dw|

23. asnaseer

does that clarify it?

24. Loser66

yup, bravo @asnaseer

25. mathcalculus

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

26. mathcalculus

@Loser66 not helping. bye

27. Loser66

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

28. asnaseer

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

29. dumbcow

@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

ooooh okay. thank you so much @dumbcow

31. mathcalculus

@asnaseer thank you!

32. asnaseer

yw :)

33. mathcalculus

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

34. mathcalculus

im not sure what to do after:?

35. asnaseer

|dw:1375032333583:dw|

36. asnaseer

the terms with (8-t) cancel out

37. asnaseer

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

38. mathcalculus

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

39. mathcalculus

or cross multiply*

40. asnaseer

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

but can we or no?

42. mathcalculus

i have not used the l hospitals rule yet.

43. asnaseer

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

yes i see that

45. asnaseer

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

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

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

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

49. asnaseer

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

50. asnaseer

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

51. asnaseer

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

52. asnaseer

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

53. asnaseer

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

54. asnaseer

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

55. asnaseer

now we can evaluate the limit as t->2

56. asnaseer

hope that makes sense?

57. mathcalculus

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

58. mathcalculus

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

59. mathcalculus

im left with 1/8t

60. asnaseer

the negative sign can just be moved up to the 1

61. mathcalculus

but not -1??

62. mathcalculus

why??

63. asnaseer

|dw:1375033388098:dw|

64. asnaseer

all 3 forms are equivalent

65. mathcalculus

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

66. mathcalculus

oay i understand that

67. asnaseer

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

68. mathcalculus

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

69. asnaseer

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

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

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

72. asnaseer

or: |dw:1375033674334:dw|

73. mathcalculus

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

74. asnaseer

75. mathcalculus

so how do i do this after?

76. asnaseer

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

77. asnaseer

|dw:1375033741276:dw|

78. mathcalculus

okay i understand.

79. asnaseer

great! :)

80. mathcalculus

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

yes :)

82. mathcalculus

thanks! =]

83. asnaseer

yw :)

84. mathcalculus

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

85. asnaseer

sure - just post it as a new question please

86. mathcalculus

87. mathcalculus

k

88. asnaseer

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