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

- anonymous

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

- schrodinger

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- alexwee123

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

- 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.

- anonymous

can you show me?

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## More answers

- anonymous

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

- anonymous

@kropot72

- 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.

- anonymous

is there another way to do this?

- anonymous

besides approaching it to the closest number?

- anonymous

not the table method.

- anonymous

@kropot72

- 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.

- anonymous

@kropot why did you multiply it by -1?

- anonymous

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

- anonymous

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

- anonymous

you were typing? lol

- Loser66

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

- anonymous

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)

- Loser66

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

- 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

- anonymous

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

- Loser66

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

- asnaseer

|dw:1375031598300:dw|

- asnaseer

does that clarify it?

- Loser66

yup, bravo @asnaseer

- anonymous

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

- anonymous

@Loser66 not helping. bye

- Loser66

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

- asnaseer

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

- 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

- anonymous

ooooh okay. thank you so much @dumbcow

- anonymous

@asnaseer thank you!

- asnaseer

yw :)

- anonymous

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

- anonymous

im not sure what to do after:?

- asnaseer

|dw:1375032333583:dw|

- asnaseer

the terms with (8-t) cancel out

- asnaseer

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

- anonymous

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

- anonymous

or cross multiply*

- 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)

- anonymous

but can we or no?

- anonymous

i have not used the l hospitals rule yet.

- 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

- anonymous

yes i see that

- 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

- anonymous

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?

- 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.

- asnaseer

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

- asnaseer

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

- asnaseer

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

- asnaseer

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

- asnaseer

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

- asnaseer

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

- asnaseer

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

- asnaseer

now we can evaluate the limit as t->2

- asnaseer

hope that makes sense?

- anonymous

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

- anonymous

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

- anonymous

im left with 1/8t

- asnaseer

the negative sign can just be moved up to the 1

- anonymous

but not -1??

- anonymous

why??

- asnaseer

|dw:1375033388098:dw|

- asnaseer

all 3 forms are equivalent

- anonymous

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

- anonymous

oay i understand that

- asnaseer

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

- anonymous

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

- 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.

- 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?

- asnaseer

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

- asnaseer

or:
|dw:1375033674334:dw|

- anonymous

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

- asnaseer

lead to the same simplification

- anonymous

so how do i do this after?

- asnaseer

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

- asnaseer

|dw:1375033741276:dw|

- anonymous

okay i understand.

- asnaseer

great! :)

- anonymous

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??

- asnaseer

yes :)

- anonymous

thanks! =]

- asnaseer

yw :)

- anonymous

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

- asnaseer

sure - just post it as a new question please

- anonymous

##### 1 Attachment

- anonymous

k

- asnaseer

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

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