Question

Find the limit, if it exists. (If an answer does not exist, enter DNE.)

lim sqroot t+ t^2/6t − t^2
t→inf

Answer 1

\[\lim_{t\to\infty}\frac{\sqrt{t+t^2}}{6t-t^2}\]is this right?

Answer 2

the sqaure root is on for the t

Answer 3

\[\lim_{t\to\infty}\frac{\sqrt t+t^2}{6t-t^2}\]

Answer 4

yeyeyeey

Answer 5

divide top and bottom by t^2

Answer 6

\[\frac{ \sqrt{t} }{ 6t }\]

Answer 7

?

Answer 8

divide each term by t^2

Answer 9

i'm oo confused can you show me

Answer 10

\[\lim_{t\to\infty}\frac{\frac1{t^2}(\sqrt t+t^2)}{\frac1{t^2}(6t-t^2)}\]

Answer 11

distribute

Answer 12

whats is the answer

Answer 13

this site is about teaching, not handing out answers
can you not simplify the numerator and denominator? this is algebra, you should know this.

Answer 14

sqroot(t)/t^2

Answer 15

6t;t^2

Answer 16

what about the t^2's on the right?

Answer 17

what is\[\frac1{t^2}(\sqrt t+t^2)\]?

Answer 18

doesn't the t^2/t^2 cancels out

Answer 19

cancels to what?

Answer 20

1

Answer 21

yes, but you never wrote the 1

Answer 22

ohhh

Answer 23

(sqroot (t)+1)/t^2/(6t-1)/t^2

Answer 24

those 1's should not be over t^2, they cancel like you said

Answer 25

just do the top\[\frac1{t^2}(\sqrt t+t^2)\]

Answer 26

okaay

Answer 27

sqroot (t)/t^2 +1/ 6t/t^2 +1

Answer 28

yes, and what is sqrt(t)/t^2 simplified?

Answer 29

i don"t know

Answer 30

\[\sqrt[b]{x^a}=x^{a/b}\]so\[\sqrt t=t^{1/2}\]and\[\frac{t^a}{t^b}=t^{a-b}\]use those two rules together

Answer 31

okaay thambi

Answer 32

3root (t^2)