Find the limit, if it exists. (If an answer does not exist, enter DNE.) lim sqroot t+ t^2/6t − t^2 t→inf

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Find the limit, if it exists. (If an answer does not exist, enter DNE.) lim sqroot t+ t^2/6t − t^2 t→inf

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

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Other answers:

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
this site is about teaching, not handing out answers can you not simplify the numerator and denominator? this is algebra, you should know this.
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?
1
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
\[\sqrt[b]{x^a}=x^{a/b}\]so\[\sqrt t=t^{1/2}\]and\[\frac{t^a}{t^b}=t^{a-b}\]use those two rules together
okaay thambi
3root (t^2)

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