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zned6559

  • one year ago

Calculate lim t-> negative infinity ((2t^2 - t -2)^1/2) / (2t+1)

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  1. amistre64
    • one year ago
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    x = sqrt(x^2) ... might be useful

  2. zned6559
    • one year ago
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    I know the answer is -1/sqrt 2 but I don't know how to get there

  3. amistre64
    • one year ago
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    ^(1/2) is the same as sqrt() id put them both under the sqrt and compare them

  4. amistre64
    • one year ago
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    \[\frac{\sqrt{2t^2-t-1}}{\sqrt{(2t+1)^2}}\]

  5. zned6559
    • one year ago
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    the bottom is suppose to be 2t+1 no sqrt

  6. amistre64
    • one year ago
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    what does sqrt(x^2) reduce to?

  7. zned6559
    • one year ago
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    x

  8. amistre64
    • one year ago
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    then sqrt((2t+1)^2) = 2t+1

  9. zned6559
    • one year ago
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    \[\sqrt{2t ^{2}-t-2}\div2t+1\]

  10. amistre64
    • one year ago
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    you are not seeing the obvious ... \[\sqrt{2t ^{2}-t-2}\div2t+1\] \[\sqrt{2t ^{2}-t-2}\div \sqrt{(2t+1)^2}\]

  11. amistre64
    • one year ago
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    what is the expansion of (2t+1)^2 ?

  12. zned6559
    • one year ago
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    \[4t ^{2}+4t+1\]

  13. amistre64
    • one year ago
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    then we have \[\sqrt{\frac{2t^2-t-2}{4t^2+4t+1}}\]

  14. amistre64
    • one year ago
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    what do we know about polynomials and their limits?

  15. zned6559
    • one year ago
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    not sure

  16. amistre64
    • one year ago
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    then you need to cover some of your basic rules .... not sure what theyve covered tho

  17. amistre64
    • one year ago
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    polys of the same degree, limit to their leading coeffs

  18. zned6559
    • one year ago
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    I got it now!

  19. amistre64
    • one year ago
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    the only thing to watch for is the sign, by squaring the bottom we altered the sign of it inthe process.

  20. amistre64
    • one year ago
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    2t+1 as t to -inf is a negative value .... so we are originally going to have a negative limit in the end

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