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alexwee123Best ResponseYou've already chosen the best response.0
is this the problem? dw:1374780769038:dw
 8 months ago

kropot72Best ResponseYou've already chosen the best response.2
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.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i've tried every way but i've gotten it incorrect a couple of times..
 8 months ago

kropot72Best ResponseYou've already chosen the best response.2
To start your table by finding the value of f(7.9) just plug 7.9 into the expression for f(t) \[f(7.9)=\frac{\frac{1}{7.9}\frac{1}{8}}{7.98}=you\ can\ calculate\] 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.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
is there another way to do this?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
besides approaching it to the closest number?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
not the table method.
 8 months ago

kropot72Best ResponseYou've already chosen the best response.2
There is a quick method. \[\frac{\frac{1}{t}\frac{1}{8}}{t8}=\frac{8t}{8t}\times\frac{1}{8t}=\frac{1}{8t}\] 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.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@kropot why did you multiply it by 1?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@dumbcow could u be able to help me with this problem?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@Loser66 i got 8t/8t but then after i'm not sure how they got 1/8t
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
you were typing? lol
 8 months ago

Loser66Best ResponseYou've already chosen the best response.0
kropot72 gave you the right logic , just plug t =8 to get 1/64, done
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
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)
 8 months ago

Loser66Best ResponseYou've already chosen the best response.0
re read the stuff, you missed something, it 's not that.
 8 months ago

dumbcowBest ResponseYou've already chosen the best response.0
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
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i know, my question is why did he multiply 1/8t ?
 8 months ago

Loser66Best ResponseYou've already chosen the best response.0
ok, dumbcow 's solution is ok, too. but kropot 72 's solution is ok, too. Yours is wrong, hehe
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
dw:1375031598300:dw
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
does that clarify it?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@asnaseer yes:) just need to know why we multiply 1/8t?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@Loser66 not helping. bye
 8 months ago

Loser66Best ResponseYou've already chosen the best response.0
ok, just because you tag me, I leave now, thanks for sending me away
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
@kropot72 made use of the fact that:\[\frac{1}{t8}=\frac{1}{8t}\]
 8 months ago

dumbcowBest ResponseYou've already chosen the best response.0
@mathcalculus , its because the denominator was "t8" the rule when dividing is multiply by reciprocal ... which turns it into 1/t8 then they factored out a neg
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
ooooh okay. thank you so much @dumbcow
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@asnaseer thank you!
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
hold oni just multiplied 8t/8t by 1/t8 and i got this: 8t/8t^264 @asnaseer
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
im not sure what to do after:?
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
dw:1375032333583:dw
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
the terms with (8t) cancel out
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
remember:\[t8=(8t)\]therefore:\[\frac{1}{t8}=\frac{1}{(8t)}=\frac{1}{8t}\]
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
yeah but i can't see that when im doing this problem alone.. im thinking i multiply across.. is that possible?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
or cross multiply*
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
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)
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
but can we or no?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i have not used the l hospitals rule yet.
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
if you use the first method, then you need to cancel out the common factors in the fraction, i.e. cancel out the (8t) from the numerator and the denominator
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
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
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i know, but all i see is that i can cross multiply but how do you know that the t8 needed to be factored?
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
it was just a /hunch/  it was a good candidate to try and cancel out as it tends to zero as t approaches 8.
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
and we want to avoid zeros (especially in the denominators) when evaluating limits
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
e.g. find the limit as t tends to 2 of:\[\frac{t^22t}{2t4}\]
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
if you just stick t=2 into this, you will get 0/0 which is undefined
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
so first we try to factorise it and see if we get anything interesting
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
\[\frac{t^22t}{2t4}=\frac{t(t2)}{2(t2)}\]
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
now we notice a (t2) can be cancelled out
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
\[\frac{t^22t}{2t4}=\frac{t\cancel{(t2)}}{2\cancel{(t2)}}=\frac{t}{2}\]
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
now we can evaluate the limit as t>2
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
hope that makes sense?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
okay so i did this now: 8t/8t * 1/ (8t)
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i cancel the 8t... but what about the negative sign outside of it???
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
im left with 1/8t
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
the negative sign can just be moved up to the 1
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
dw:1375033388098:dw
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
all 3 forms are equivalent
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i need rules, not simply "oh because we can." "just guessing" "that works"
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
oay i understand that
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
the rules here are to look for common factors that can be cancelled out
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
so when i asked u if we can multiply across, can we or not?
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
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.
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
you CAN multiply across but it won't simplify the fraction. why wouldn't you want to cancel out the common terms BEFORE multiplying?
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
e.g.: dw:1375033624859:dw
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
or: dw:1375033674334:dw
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
okay from here:8t/8t^264t
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
lead to the same simplification
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
so how do i do this after?
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
yes  then you will have to refactorise the denominator to cancel out the (8t)
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
dw:1375033741276:dw
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
okay i understand.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
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??
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
can we go over another problem so i can feel more confident on it? @asneer
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
sure  just post it as a new question please
 8 months ago

asnaseerBest ResponseYou've already chosen the best response.1
<<< please post as a new question in the list on the left it helps others learn as well :)
 8 months ago
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