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

Mathematics
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is this the problem? |dw:1374780769038:dw|
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.
can you show me?

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i've tried every way but i've gotten it incorrect a couple of times..
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.
is there another way to do this?
besides approaching it to the closest number?
not the table method.
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.
@kropot why did you multiply it by -1?
@dumbcow could u be able to help me with this problem?
@Loser66 i got 8-t/8t but then after i'm not sure how they got -1/8t
you were typing? lol
kropot72 gave you the right logic , just plug t =8 to get -1/64, done
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)
re read the stuff, you missed something, it 's not that.
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
i know, my question is why did he multiply -1/8-t ?
ok, dumbcow 's solution is ok, too. but kropot 72 's solution is ok, too. Yours is wrong, hehe
|dw:1375031598300:dw|
does that clarify it?
yup, bravo @asnaseer
@asnaseer yes:) just need to know why we multiply -1/8-t?
@Loser66 not helping. bye
ok, just because you tag me, I leave now, thanks for sending me away
@kropot72 made use of the fact that:\[\frac{1}{t-8}=\frac{-1}{8-t}\]
@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
ooooh okay. thank you so much @dumbcow
@asnaseer thank you!
yw :)
hold oni just multiplied 8-t/8t by 1/t-8 and i got this: 8-t/8t^2-64 @asnaseer
im not sure what to do after:?
|dw:1375032333583:dw|
the terms with (8-t) cancel out
remember:\[t-8=-(8-t)\]therefore:\[\frac{1}{t-8}=\frac{1}{-(8-t)}=\frac{-1}{8-t}\]
yeah but i can't see that when im doing this problem alone.. im thinking i multiply across.. is that possible?
or cross multiply*
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)
but can we or no?
i have not used the l hospitals rule yet.
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
yes i see that
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
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?
it was just a /hunch/ - it was a good candidate to try and cancel out as it tends to zero as t approaches 8.
and we want to avoid zeros (especially in the denominators) when evaluating limits
e.g. find the limit as t tends to 2 of:\[\frac{t^2-2t}{2t-4}\]
if you just stick t=2 into this, you will get 0/0 which is undefined
so first we try to factorise it and see if we get anything interesting
\[\frac{t^2-2t}{2t-4}=\frac{t(t-2)}{2(t-2)}\]
now we notice a (t-2) can be cancelled out
\[\frac{t^2-2t}{2t-4}=\frac{t\cancel{(t-2)}}{2\cancel{(t-2)}}=\frac{t}{2}\]
now we can evaluate the limit as t->2
hope that makes sense?
okay so i did this now: 8-t/8t * 1/- (8-t)
i cancel the 8-t... but what about the negative sign outside of it???
im left with 1/8t
the negative sign can just be moved up to the 1
but not -1??
why??
|dw:1375033388098:dw|
all 3 forms are equivalent
i need rules, not simply "oh because we can." "just guessing" "that works"
oay i understand that
the rules here are to look for common factors that can be cancelled out
so when i asked u if we can multiply across, can we or not?
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.
you CAN multiply across but it won't simplify the fraction. why wouldn't you want to cancel out the common terms BEFORE multiplying?
e.g.: |dw:1375033624859:dw|
or: |dw:1375033674334:dw|
okay from here:8-t/8t^2-64t
lead to the same simplification
so how do i do this after?
yes - then you will have to re-factorise the denominator to cancel out the (8-t)
|dw:1375033741276:dw|
okay i understand.
great! :)
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??
yes :)
thanks! =]
yw :)
can we go over another problem so i can feel more confident on it? @asneer
sure - just post it as a new question please
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k
<<<--- please post as a new question in the list on the left it helps others learn as well :)

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