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El_Arrow

  • one year ago

don't understand this limit problem please help

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  1. El_Arrow
    • one year ago
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  2. El_Arrow
    • one year ago
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    i dont understand why she did that

  3. anonymous
    • one year ago
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    divide top and bottom by \(n^2\) so you can more easily see which terms dominate

  4. El_Arrow
    • one year ago
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    but if you do that in the bottom for example n^3/n^2 you get n not 1/n

  5. anonymous
    • one year ago
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    $$\frac{3n^2+2n}{\sqrt{n^3+n^2+1}}\cdot\frac{1/n^2}{1/n^2}=\frac{3n^2+2n}{\sqrt{n^3+n^2+1}}\cdot\frac{1/n^2}{\sqrt{1/n^4}}=\frac{3+2/n}{\sqrt{1/n+1/n^2+1/n^4}}$$ so you can clearly

  6. El_Arrow
    • one year ago
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    oh so you square it when there is a square root?

  7. anonymous
    • one year ago
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    Actually he did that..

  8. anonymous
    • one year ago
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    Oh sorry, that was someone else you did that..

  9. anonymous
    • one year ago
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    see that in the limit as \(n\to\infty\), the square root tends to \(0\) while the top tends to \(3\); this tells us the terms 'blow up' as \(n\) grows larger and larger since the denominator gets ever smaller

  10. anonymous
    • one year ago
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    \[\sqrt{\frac{1}{x^2}} = \frac{1}{x} = \sqrt{\frac{1}{x^2}}\]

  11. anonymous
    • one year ago
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    *who..

  12. anonymous
    • one year ago
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    When you take x inside squares, it becomes \(x^2\)..

  13. anonymous
    • one year ago
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    *square root brackets..

  14. El_Arrow
    • one year ago
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    oh okay i see

  15. El_Arrow
    • one year ago
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    i have one more question

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  16. El_Arrow
    • one year ago
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    where did the 8 in this problem come from?

  17. anonymous
    • one year ago
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    the distance between the vertex and focus (\(-4\)) is equal to the distance between the vertex and directrix

  18. El_Arrow
    • one year ago
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    so you are add -4+-4?

  19. anonymous
    • one year ago
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    |dw:1437367323144:dw|

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spraguer (Moderator)
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