mathcalculus
HELP: Evaluate the limit:
lim ((1/t)-(1/8))/ (t-8) =
t->8
Delete
Share
This Question is Closed
alexwee123
Best Response
You've already chosen the best response.
0
is this the problem?
|dw:1374780769038:dw|
kropot72
Best Response
You'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.
mathcalculus
Best Response
You've already chosen the best response.
0
can you show me?
mathcalculus
Best Response
You've already chosen the best response.
0
i've tried every way but i've gotten it incorrect a couple of times..
mathcalculus
Best Response
You've already chosen the best response.
0
@kropot72
kropot72
Best Response
You'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)
\[f(7.9)=\frac{\frac{1}{7.9}-\frac{1}{8}}{7.9-8}=you\ can\ calculate\]
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.
mathcalculus
Best Response
You've already chosen the best response.
0
is there another way to do this?
mathcalculus
Best Response
You've already chosen the best response.
0
besides approaching it to the closest number?
mathcalculus
Best Response
You've already chosen the best response.
0
not the table method.
mathcalculus
Best Response
You've already chosen the best response.
0
@kropot72
kropot72
Best Response
You've already chosen the best response.
2
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.
mathcalculus
Best Response
You've already chosen the best response.
0
@kropot why did you multiply it by -1?
mathcalculus
Best Response
You've already chosen the best response.
0
@dumbcow could u be able to help me with this problem?
mathcalculus
Best Response
You've already chosen the best response.
0
@Loser66 i got 8-t/8t but then after i'm not sure how they got -1/8t
mathcalculus
Best Response
You've already chosen the best response.
0
you were typing? lol
Loser66
Best Response
You've already chosen the best response.
0
kropot72 gave you the right logic , just plug t =8 to get -1/64, done
mathcalculus
Best Response
You'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)
Loser66
Best Response
You've already chosen the best response.
0
re read the stuff, you missed something, it 's not that.
dumbcow
Best Response
You'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
mathcalculus
Best Response
You've already chosen the best response.
0
i know, my question is why did he multiply -1/8-t ?
Loser66
Best Response
You'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
asnaseer
Best Response
You've already chosen the best response.
1
|dw:1375031598300:dw|
asnaseer
Best Response
You've already chosen the best response.
1
does that clarify it?
Loser66
Best Response
You've already chosen the best response.
0
yup, bravo @asnaseer
mathcalculus
Best Response
You've already chosen the best response.
0
@asnaseer yes:) just need to know why we multiply -1/8-t?
mathcalculus
Best Response
You've already chosen the best response.
0
@Loser66 not helping. bye
Loser66
Best Response
You've already chosen the best response.
0
ok, just because you tag me, I leave now, thanks for sending me away
asnaseer
Best Response
You've already chosen the best response.
1
@kropot72 made use of the fact that:\[\frac{1}{t-8}=\frac{-1}{8-t}\]
dumbcow
Best Response
You've already chosen the best response.
0
@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
mathcalculus
Best Response
You've already chosen the best response.
0
ooooh okay. thank you so much @dumbcow
mathcalculus
Best Response
You've already chosen the best response.
0
@asnaseer thank you!
asnaseer
Best Response
You've already chosen the best response.
1
yw :)
mathcalculus
Best Response
You've already chosen the best response.
0
hold oni just multiplied 8-t/8t by 1/t-8 and i got this: 8-t/8t^2-64 @asnaseer
mathcalculus
Best Response
You've already chosen the best response.
0
im not sure what to do after:?
asnaseer
Best Response
You've already chosen the best response.
1
|dw:1375032333583:dw|
asnaseer
Best Response
You've already chosen the best response.
1
the terms with (8-t) cancel out
asnaseer
Best Response
You've already chosen the best response.
1
remember:\[t-8=-(8-t)\]therefore:\[\frac{1}{t-8}=\frac{1}{-(8-t)}=\frac{-1}{8-t}\]
mathcalculus
Best Response
You'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?
mathcalculus
Best Response
You've already chosen the best response.
0
or cross multiply*
asnaseer
Best Response
You'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)
mathcalculus
Best Response
You've already chosen the best response.
0
but can we or no?
mathcalculus
Best Response
You've already chosen the best response.
0
i have not used the l hospitals rule yet.
asnaseer
Best Response
You'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 (8-t) from the numerator and the denominator
mathcalculus
Best Response
You've already chosen the best response.
0
yes i see that
asnaseer
Best Response
You'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
mathcalculus
Best Response
You'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 t-8 needed to be factored?
asnaseer
Best Response
You'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.
asnaseer
Best Response
You've already chosen the best response.
1
and we want to avoid zeros (especially in the denominators) when evaluating limits
asnaseer
Best Response
You've already chosen the best response.
1
e.g. find the limit as t tends to 2 of:\[\frac{t^2-2t}{2t-4}\]
asnaseer
Best Response
You've already chosen the best response.
1
if you just stick t=2 into this, you will get 0/0 which is undefined
asnaseer
Best Response
You've already chosen the best response.
1
so first we try to factorise it and see if we get anything interesting
asnaseer
Best Response
You've already chosen the best response.
1
\[\frac{t^2-2t}{2t-4}=\frac{t(t-2)}{2(t-2)}\]
asnaseer
Best Response
You've already chosen the best response.
1
now we notice a (t-2) can be cancelled out
asnaseer
Best Response
You've already chosen the best response.
1
\[\frac{t^2-2t}{2t-4}=\frac{t\cancel{(t-2)}}{2\cancel{(t-2)}}=\frac{t}{2}\]
asnaseer
Best Response
You've already chosen the best response.
1
now we can evaluate the limit as t->2
asnaseer
Best Response
You've already chosen the best response.
1
hope that makes sense?
mathcalculus
Best Response
You've already chosen the best response.
0
okay so i did this now:
8-t/8t * 1/- (8-t)
mathcalculus
Best Response
You've already chosen the best response.
0
i cancel the 8-t... but what about the negative sign outside of it???
mathcalculus
Best Response
You've already chosen the best response.
0
im left with 1/8t
asnaseer
Best Response
You've already chosen the best response.
1
the negative sign can just be moved up to the 1
mathcalculus
Best Response
You've already chosen the best response.
0
but not -1??
mathcalculus
Best Response
You've already chosen the best response.
0
why??
asnaseer
Best Response
You've already chosen the best response.
1
|dw:1375033388098:dw|
asnaseer
Best Response
You've already chosen the best response.
1
all 3 forms are equivalent
mathcalculus
Best Response
You've already chosen the best response.
0
i need rules, not simply "oh because we can." "just guessing" "that works"
mathcalculus
Best Response
You've already chosen the best response.
0
oay i understand that
asnaseer
Best Response
You've already chosen the best response.
1
the rules here are to look for common factors that can be cancelled out
mathcalculus
Best Response
You've already chosen the best response.
0
so when i asked u if we can multiply across, can we or not?
asnaseer
Best Response
You'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.
asnaseer
Best Response
You'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?
asnaseer
Best Response
You've already chosen the best response.
1
e.g.:
|dw:1375033624859:dw|
asnaseer
Best Response
You've already chosen the best response.
1
or:
|dw:1375033674334:dw|
mathcalculus
Best Response
You've already chosen the best response.
0
okay from here:8-t/8t^2-64t
asnaseer
Best Response
You've already chosen the best response.
1
lead to the same simplification
mathcalculus
Best Response
You've already chosen the best response.
0
so how do i do this after?
asnaseer
Best Response
You've already chosen the best response.
1
yes - then you will have to re-factorise the denominator to cancel out the (8-t)
asnaseer
Best Response
You've already chosen the best response.
1
|dw:1375033741276:dw|
mathcalculus
Best Response
You've already chosen the best response.
0
okay i understand.
asnaseer
Best Response
You've already chosen the best response.
1
great! :)
mathcalculus
Best Response
You'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??
asnaseer
Best Response
You've already chosen the best response.
1
yes :)
mathcalculus
Best Response
You've already chosen the best response.
0
thanks! =]
asnaseer
Best Response
You've already chosen the best response.
1
yw :)
mathcalculus
Best Response
You've already chosen the best response.
0
can we go over another problem so i can feel more confident on it? @asneer
asnaseer
Best Response
You've already chosen the best response.
1
sure - just post it as a new question please
mathcalculus
Best Response
You've already chosen the best response.
0
mathcalculus
Best Response
You've already chosen the best response.
0
k
asnaseer
Best Response
You'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 :)