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

the sqaure root is on for the t

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

yeyeyeey

divide top and bottom by t^2

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

divide each term by t^2

i'm oo confused can you show me

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

distribute

whats is the answer

sqroot(t)/t^2

6t;t^2

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

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

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

cancels to what?

yes, but you never wrote the 1

ohhh

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

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

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

okaay

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

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

i don"t know

okaay thambi

3root (t^2)