At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

is this the problem?
|dw:1374780769038:dw|

can you show me?

i've tried every way but i've gotten it incorrect a couple of times..

is there another way to do this?

besides approaching it to the closest number?

not the table method.

you were typing? lol

kropot72 gave you the right logic , just plug t =8 to get -1/64, done

re read the stuff, you missed something, it 's not that.

i know, my question is why did he multiply -1/8-t ?

ok, dumbcow 's solution is ok, too. but kropot 72 's solution is ok, too. Yours is wrong, hehe

|dw:1375031598300:dw|

does that clarify it?

ok, just because you tag me, I leave now, thanks for sending me away

yw :)

im not sure what to do after:?

|dw:1375032333583:dw|

the terms with (8-t) cancel out

remember:\[t-8=-(8-t)\]therefore:\[\frac{1}{t-8}=\frac{1}{-(8-t)}=\frac{-1}{8-t}\]

or cross multiply*

but can we or no?

i have not used the l hospitals rule yet.

yes i see that

and we want to avoid zeros (especially in the denominators) when evaluating limits

e.g. find the limit as t tends to 2 of:\[\frac{t^2-2t}{2t-4}\]

if you just stick t=2 into this, you will get 0/0 which is undefined

so first we try to factorise it and see if we get anything interesting

\[\frac{t^2-2t}{2t-4}=\frac{t(t-2)}{2(t-2)}\]

now we notice a (t-2) can be cancelled out

\[\frac{t^2-2t}{2t-4}=\frac{t\cancel{(t-2)}}{2\cancel{(t-2)}}=\frac{t}{2}\]

now we can evaluate the limit as t->2

hope that makes sense?

okay so i did this now:
8-t/8t * 1/- (8-t)

i cancel the 8-t... but what about the negative sign outside of it???

im left with 1/8t

the negative sign can just be moved up to the 1

but not -1??

why??

|dw:1375033388098:dw|

all 3 forms are equivalent

i need rules, not simply "oh because we can." "just guessing" "that works"

oay i understand that

the rules here are to look for common factors that can be cancelled out

so when i asked u if we can multiply across, can we or not?

e.g.:
|dw:1375033624859:dw|

or:
|dw:1375033674334:dw|

okay from here:8-t/8t^2-64t

lead to the same simplification

so how do i do this after?

yes - then you will have to re-factorise the denominator to cancel out the (8-t)

|dw:1375033741276:dw|

okay i understand.

great! :)

yes :)

thanks! =]

yw :)

sure - just post it as a new question please

<<<--- please post as a new question in the list on the left
it helps others learn as well :)