Question

Limit ( t^3 + t^2 - 5t +3 / t^3 - 3t + 2)
x tends to -1

Answer 1

you mean t tends to -1?

Answer 2

yeah

Answer 3

tell me tano plzzzz

Answer 4

did you try replacing t for -1?

Answer 5

yeah but it got infinity .....

Answer 6

i knw its answer is 4/3 but dont knw how to solve

Answer 7

the answer is 2

Answer 8

if you replace t for -1 , you get 8/4 \[=2\]

Answer 9

\[{(-1)^3 +(-1)^2-5(-1)+ 3\over (-1)^3-3(-1)+2} = {-1 + 1 + 5 + 3\over -1 + 3 + 2} = {8\over 4} = 2\]

Answer 10

wrong dear

Answer 11

(-1)^2 = 1 not -1 :(

Answer 12

Pretty sure that's what I wrote..

Answer 13

sry

Answer 14

hey

Answer 15

polpak srry the limit was t tends to 1

Answer 16

limit is 1 thn tell me solution

Answer 17

??????

Answer 18

Oh.. You should say what you mean then. ;p

You need to take the derivative of the top and bottom and solve for that limit. By l'Hopital's rule it will be the same.

Answer 19

ok but its computing limit by simple method not by L hopital rule
is there is any other way ??

Answer 20

Oh, sure you can do that too.

Answer 21

hmm tell me?

Answer 22

w8ing

Answer 23

Wait no. blah

Answer 24

wrong
denominator

Answer 25

You have to use l'Hopital.

Answer 26

:( ok

Answer 27

Yeah, if it were going to 0 you would be fine with just dividing.

Answer 28

But it's not, so you cannot ;p

Answer 29

i was tried all this step but .... i thng L hopital rule is only way

Answer 30

thnx buddy 4 ur time