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

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\[\lim_{t\to\infty}\frac{\sqrt{t+t^2}}{6tt^2}\]is this right?
 one year ago

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the sqaure root is on for the t
 one year ago

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

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divide top and bottom by t^2
 one year ago

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\[\frac{ \sqrt{t} }{ 6t }\]
 one year ago

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divide each term by t^2
 one year ago

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i'm oo confused can you show me
 one year ago

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

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distribute
 one year ago

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whats is the answer
 one year ago

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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.
 one year ago

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what about the t^2's on the right?
 one year ago

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

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doesn't the t^2/t^2 cancels out
 one year ago

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cancels to what?
 one year ago

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yes, but you never wrote the 1
 one year ago

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(sqroot (t)+1)/t^2/(6t1)/t^2
 one year ago

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

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

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sqroot (t)/t^2 +1/ 6t/t^2 +1
 one year ago

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

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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
 one year ago
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