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mathcalculus
 2 years ago
HELP: Evaluate the limit:
lim ((1/t)(1/8))/ (t8) =
t>8
mathcalculus
 2 years ago
HELP: Evaluate the limit: lim ((1/t)(1/8))/ (t8) = t>8

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alexwee123
 2 years ago
Best ResponseYou've already chosen the best response.0is this the problem? dw:1374780769038:dw

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2To 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.

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0i've tried every way but i've gotten it incorrect a couple of times..

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2To 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.

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0is there another way to do this?

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0besides approaching it to the closest number?

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0not the table method.

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.2There 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.

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0@kropot why did you multiply it by 1?

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0@dumbcow could u be able to help me with this problem?

mathcalculus
 2 years ago
Best 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

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0you were typing? lol

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

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0yeah 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
 2 years ago
Best ResponseYou've already chosen the best response.0re read the stuff, you missed something, it 's not that.

dumbcow
 2 years ago
Best ResponseYou've already chosen the best response.0another 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

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

Loser66
 2 years ago
Best ResponseYou've already chosen the best response.0ok, dumbcow 's solution is ok, too. but kropot 72 's solution is ok, too. Yours is wrong, hehe

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

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0@Loser66 not helping. bye

Loser66
 2 years ago
Best ResponseYou've already chosen the best response.0ok, just because you tag me, I leave now, thanks for sending me away

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

dumbcow
 2 years ago
Best 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

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0ooooh okay. thank you so much @dumbcow

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0@asnaseer thank you!

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

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0im not sure what to do after:?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1the terms with (8t) cancel out

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

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

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0or cross multiply*

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1in 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)

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0i have not used the l hospitals rule yet.

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1if 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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1if 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

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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1and we want to avoid zeros (especially in the denominators) when evaluating limits

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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1so first we try to factorise it and see if we get anything interesting

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1now we notice a (t2) can be cancelled out

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1now we can evaluate the limit as t>2

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1hope that makes sense?

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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1the negative sign can just be moved up to the 1

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1all 3 forms are equivalent

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0i need rules, not simply "oh because we can." "just guessing" "that works"

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0oay i understand that

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1the rules here are to look for common factors that can be cancelled out

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0so when i asked u if we can multiply across, can we or not?

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1e.g.: dw:1375033624859:dw

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1or: dw:1375033674334:dw

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0okay from here:8t/8t^264t

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1lead to the same simplification

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0so how do i do this after?

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

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0okay i understand.

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0now 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??

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1sure  just post it as a new question please

asnaseer
 2 years ago
Best 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 :)
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