v.s
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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\[\lim_{t\to\infty}\frac{\sqrt{t+t^2}}{6tt^2}\]is this right?

v.s
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the sqaure root is on for the t

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

v.s
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yeyeyeey

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divide top and bottom by t^2

v.s
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\[\frac{ \sqrt{t} }{ 6t }\]

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?

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divide each term by t^2

v.s
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i'm oo confused can you show me

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

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distribute

v.s
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whats is the answer

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

v.s
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sqroot(t)/t^2

v.s
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6t;t^2

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

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

v.s
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doesn't the t^2/t^2 cancels out

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cancels to what?

v.s
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1

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yes, but you never wrote the 1

v.s
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ohhh

v.s
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(sqroot (t)+1)/t^2/(6t1)/t^2

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

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

v.s
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okaay

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

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

v.s
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i don"t know

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

v.s
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okaay thambi

v.s
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3root (t^2)