anonymous
  • anonymous
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
schrodinger
  • schrodinger
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TuringTest
  • TuringTest
\[\lim_{t\to\infty}\frac{\sqrt{t+t^2}}{6t-t^2}\]is this right?
anonymous
  • anonymous
the sqaure root is on for the t
TuringTest
  • TuringTest
\[\lim_{t\to\infty}\frac{\sqrt t+t^2}{6t-t^2}\]

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More answers

anonymous
  • anonymous
yeyeyeey
TuringTest
  • TuringTest
divide top and bottom by t^2
anonymous
  • anonymous
\[\frac{ \sqrt{t} }{ 6t }\]
TuringTest
  • TuringTest
?
TuringTest
  • TuringTest
divide each term by t^2
anonymous
  • anonymous
i'm oo confused can you show me
TuringTest
  • TuringTest
\[\lim_{t\to\infty}\frac{\frac1{t^2}(\sqrt t+t^2)}{\frac1{t^2}(6t-t^2)}\]
TuringTest
  • TuringTest
distribute
anonymous
  • anonymous
whats is the answer
TuringTest
  • TuringTest
this site is about teaching, not handing out answers can you not simplify the numerator and denominator? this is algebra, you should know this.
anonymous
  • anonymous
sqroot(t)/t^2
anonymous
  • anonymous
6t;t^2
TuringTest
  • TuringTest
what about the t^2's on the right?
TuringTest
  • TuringTest
what is\[\frac1{t^2}(\sqrt t+t^2)\]?
anonymous
  • anonymous
doesn't the t^2/t^2 cancels out
TuringTest
  • TuringTest
cancels to what?
anonymous
  • anonymous
1
TuringTest
  • TuringTest
yes, but you never wrote the 1
anonymous
  • anonymous
ohhh
anonymous
  • anonymous
(sqroot (t)+1)/t^2/(6t-1)/t^2
TuringTest
  • TuringTest
those 1's should not be over t^2, they cancel like you said
TuringTest
  • TuringTest
just do the top\[\frac1{t^2}(\sqrt t+t^2)\]
anonymous
  • anonymous
okaay
anonymous
  • anonymous
sqroot (t)/t^2 +1/ 6t/t^2 +1
TuringTest
  • TuringTest
yes, and what is sqrt(t)/t^2 simplified?
anonymous
  • anonymous
i don"t know
TuringTest
  • TuringTest
\[\sqrt[b]{x^a}=x^{a/b}\]so\[\sqrt t=t^{1/2}\]and\[\frac{t^a}{t^b}=t^{a-b}\]use those two rules together
anonymous
  • anonymous
okaay thambi
anonymous
  • anonymous
3root (t^2)

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