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v.s

  • 2 years ago

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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  1. TuringTest
    • 2 years ago
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    \[\lim_{t\to\infty}\frac{\sqrt{t+t^2}}{6t-t^2}\]is this right?

  2. v.s
    • 2 years ago
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    the sqaure root is on for the t

  3. TuringTest
    • 2 years ago
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    \[\lim_{t\to\infty}\frac{\sqrt t+t^2}{6t-t^2}\]

  4. v.s
    • 2 years ago
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    yeyeyeey

  5. TuringTest
    • 2 years ago
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    divide top and bottom by t^2

  6. v.s
    • 2 years ago
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    \[\frac{ \sqrt{t} }{ 6t }\]

  7. TuringTest
    • 2 years ago
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    ?

  8. TuringTest
    • 2 years ago
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    divide each term by t^2

  9. v.s
    • 2 years ago
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    i'm oo confused can you show me

  10. TuringTest
    • 2 years ago
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    \[\lim_{t\to\infty}\frac{\frac1{t^2}(\sqrt t+t^2)}{\frac1{t^2}(6t-t^2)}\]

  11. TuringTest
    • 2 years ago
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    distribute

  12. v.s
    • 2 years ago
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    whats is the answer

  13. TuringTest
    • 2 years 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.

  14. v.s
    • 2 years ago
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    sqroot(t)/t^2

  15. v.s
    • 2 years ago
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    6t;t^2

  16. TuringTest
    • 2 years ago
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    what about the t^2's on the right?

  17. TuringTest
    • 2 years ago
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    what is\[\frac1{t^2}(\sqrt t+t^2)\]?

  18. v.s
    • 2 years ago
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    doesn't the t^2/t^2 cancels out

  19. TuringTest
    • 2 years ago
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    cancels to what?

  20. v.s
    • 2 years ago
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    1

  21. TuringTest
    • 2 years ago
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    yes, but you never wrote the 1

  22. v.s
    • 2 years ago
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    ohhh

  23. v.s
    • 2 years ago
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    (sqroot (t)+1)/t^2/(6t-1)/t^2

  24. TuringTest
    • 2 years ago
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    those 1's should not be over t^2, they cancel like you said

  25. TuringTest
    • 2 years ago
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    just do the top\[\frac1{t^2}(\sqrt t+t^2)\]

  26. v.s
    • 2 years ago
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    okaay

  27. v.s
    • 2 years ago
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    sqroot (t)/t^2 +1/ 6t/t^2 +1

  28. TuringTest
    • 2 years ago
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    yes, and what is sqrt(t)/t^2 simplified?

  29. v.s
    • 2 years ago
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    i don"t know

  30. TuringTest
    • 2 years ago
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    \[\sqrt[b]{x^a}=x^{a/b}\]so\[\sqrt t=t^{1/2}\]and\[\frac{t^a}{t^b}=t^{a-b}\]use those two rules together

  31. v.s
    • 2 years ago
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    okaay thambi

  32. v.s
    • 2 years ago
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    3root (t^2)

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