anonymous
  • anonymous
HELP: Evaluate the limit: lim ((1/t)-(1/8))/ (t-8) = t->8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
alexwee123
  • alexwee123
is this the problem? |dw:1374780769038:dw|
kropot72
  • kropot72
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.
anonymous
  • anonymous
can you show me?

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

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