|
|
|
| Other
|
| Math
|
| Biology
|
| Business
|
| Chemistry
|
| Economics
|
| Essay Service
|
| History
|
| Language
|
| Physics
|
| z Medical Questions
|
|
|
|
 |
| Post
Answer |
What is the limit as t approaches 1 for (t^3+t^2-5t+3)/(t^3-3t+2)?
I know the answer is 1 and 1/3, but I used my calculator for that. Problem is, my upcoming test is a no calculator test. Can you please show the steps to get the answer? |
| Answer |
above problem can be written as
(t^3+3t^2-2t^2-6t+t+3)/
(t^3-2t^2+2t^2+t-4t+2)
rearranging
(t^3-2t^2+t+3t^2-6t+3)/
(t^3-2t^2+t+2t^2-4t+2)
taking common
[t(t^2-2t+1)+3(t^2-2t+1)]/
[t(t^2-2t+1)+2(t^2-2t+1)]
which is equal to
(t+3)(t^2-2t+1)/
(t+2)(t^2-2t+1)
cancelling (t^2-2t+1) since t approaches to 1 i.e. t is not equal to 1
so
(t+3)/(t+2)
putting t=1 as t approaches 1
ans is
4/3
i think u have written 1+ 1/3
i.e. 4/3 |
| Amount
of $ Requested |
$
1.00 |
|
|
|
