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mathcalculus
Group Title
HELP: Evaluate the limit:
lim ((1/t)(1/8))/ (t8) =
t>8
 one year ago
 one year ago
mathcalculus Group Title
HELP: Evaluate the limit: lim ((1/t)(1/8))/ (t8) = t>8
 one year ago
 one year ago

This Question is Closed

alexwee123 Group TitleBest ResponseYou've already chosen the best response.0
is this the problem? dw:1374780769038:dw
 one year ago

kropot72 Group TitleBest 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.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
can you show me?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i've tried every way but i've gotten it incorrect a couple of times..
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@kropot72
 one year ago

kropot72 Group TitleBest 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.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
is there another way to do this?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
besides approaching it to the closest number?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
not the table method.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@kropot72
 one year ago

kropot72 Group TitleBest 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.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@kropot why did you multiply it by 1?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@dumbcow could u be able to help me with this problem?
 one year ago

mathcalculus Group TitleBest 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
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
you were typing? lol
 one year ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
kropot72 gave you the right logic , just plug t =8 to get 1/64, done
 one year ago

mathcalculus Group TitleBest 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)
 one year ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
re read the stuff, you missed something, it 's not that.
 one year ago

dumbcow Group TitleBest 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
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i know, my question is why did he multiply 1/8t ?
 one year ago

Loser66 Group TitleBest 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
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
dw:1375031598300:dw
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
does that clarify it?
 one year ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
yup, bravo @asnaseer
 one year ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@Loser66 not helping. bye
 one year ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
ok, just because you tag me, I leave now, thanks for sending me away
 one year ago

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

dumbcow Group TitleBest 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
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
ooooh okay. thank you so much @dumbcow
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@asnaseer thank you!
 one year ago

mathcalculus Group TitleBest 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
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
im not sure what to do after:?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
dw:1375032333583:dw
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
the terms with (8t) cancel out
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
remember:\[t8=(8t)\]therefore:\[\frac{1}{t8}=\frac{1}{(8t)}=\frac{1}{8t}\]
 one year ago

mathcalculus Group TitleBest 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?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
or cross multiply*
 one year ago

asnaseer Group TitleBest 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)
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
but can we or no?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i have not used the l hospitals rule yet.
 one year ago

asnaseer Group TitleBest 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
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yes i see that
 one year ago

asnaseer Group TitleBest 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
 one year ago

mathcalculus Group TitleBest 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?
 one year ago

asnaseer Group TitleBest 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.
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
and we want to avoid zeros (especially in the denominators) when evaluating limits
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
e.g. find the limit as t tends to 2 of:\[\frac{t^22t}{2t4}\]
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
if you just stick t=2 into this, you will get 0/0 which is undefined
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
so first we try to factorise it and see if we get anything interesting
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{t^22t}{2t4}=\frac{t(t2)}{2(t2)}\]
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
now we notice a (t2) can be cancelled out
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{t^22t}{2t4}=\frac{t\cancel{(t2)}}{2\cancel{(t2)}}=\frac{t}{2}\]
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
now we can evaluate the limit as t>2
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
hope that makes sense?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
okay so i did this now: 8t/8t * 1/ (8t)
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i cancel the 8t... but what about the negative sign outside of it???
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
im left with 1/8t
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
the negative sign can just be moved up to the 1
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
but not 1??
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
why??
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
dw:1375033388098:dw
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
all 3 forms are equivalent
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i need rules, not simply "oh because we can." "just guessing" "that works"
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oay i understand that
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
the rules here are to look for common factors that can be cancelled out
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
so when i asked u if we can multiply across, can we or not?
 one year ago

asnaseer Group TitleBest 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.
 one year ago

asnaseer Group TitleBest 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?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
e.g.: dw:1375033624859:dw
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
or: dw:1375033674334:dw
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
okay from here:8t/8t^264t
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
lead to the same simplification
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
so how do i do this after?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
yes  then you will have to refactorise the denominator to cancel out the (8t)
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
dw:1375033741276:dw
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
okay i understand.
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
great! :)
 one year ago

mathcalculus Group TitleBest 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??
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
thanks! =]
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
can we go over another problem so i can feel more confident on it? @asneer
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
sure  just post it as a new question please
 one year ago

asnaseer Group TitleBest 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 :)
 one year ago
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