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CoconutJJ

  • one year ago

LIM{x -> inf} [sqrt( (X^2) - 3x) - x]

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  1. CoconutJJ
    • one year ago
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    \[\lim_{x \rightarrow \infty}(\sqrt{x^2-3x}-x)\]

  2. Jhannybean
    • one year ago
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    \[\large \lim_{x\rightarrow \infty} \sqrt{x^2-3x}-x\] simplify this by multiplying it by its conjugate?

  3. CoconutJJ
    • one year ago
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    I tried that, but then the denominator becomes a problem... :(

  4. Jhannybean
    • one year ago
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    \[\sqrt{x^2-3x}-x \cdot \frac{\sqrt{x^2-3x}+x}{\sqrt{x^2-3x}+x}\]\[=\frac{(x^2-3x)+x\sqrt{x^2-3x}-x\sqrt{x^2-3x}-x^2}{\sqrt{x^2-3x}+x}\]\[=\frac{-3x}{\sqrt{x^2-3x}+x}\]

  5. Jhannybean
    • one year ago
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    Take the derivative of \(\sqrt{x^2-3x}+x\) and see what you get first.

  6. Jhannybean
    • one year ago
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    This limit should come out to \(-\dfrac{3}{2}\)

  7. CoconutJJ
    • one year ago
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    yea.. I can use l'hopital's rule, however I'm only on chapter 1 and the teacher won't allow that

  8. Jhannybean
    • one year ago
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    Ah, =_= ok.

  9. Jhannybean
    • one year ago
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    What if we... divided by the smallest power. \[\large \lim_{x\rightarrow \infty} \frac{-\dfrac{3x}{x}}{\sqrt{\dfrac{x^2}{x} - \dfrac{3x}{x}} +\dfrac{x}{x}}\]

  10. CoconutJJ
    • one year ago
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    however, when you move x into the radical it becomes x^2

  11. Empty
    • one year ago
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    substitute in: \[x^2-3x=y^2\] solve for x to plug it in everywhere and you get: \[\lim_{y \to \infty} \sqrt{y^2}-\frac{3+\sqrt{9+4y^2}}{2}\] \[\frac{-3}{2}+\lim_{y \to \infty}y-\frac{\sqrt{9+4y^2}}{2} \] Not really sure if this is right path to go or not, but we're getting that -3/2 that the great wolfram alpha gives.

  12. CoconutJJ
    • one year ago
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    Hmm I can't seem to find the correct answer with substitution either.. I just end up with the same equation again

  13. Jhannybean
    • one year ago
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    Oh,I was forgetting a fundamental rule... \[\sqrt{x^2} = |x| = x ~~~\text{as x goes towards infinity}\]Therefore \[\large \lim_{x\rightarrow \infty} \frac{-\dfrac{3x}{x}}{\sqrt{\dfrac{x^2}{x^2} - \dfrac{3x}{x^2}} +\dfrac{x}{x}}\]\[\large \lim_{\rightarrow \infty}\frac{-3}{\sqrt{1-\dfrac{3}{x}}+1}\]\[\large \frac{-3}{\sqrt{1}+1} = -\frac{3}{2}\]

  14. IrishBoy123
    • one year ago
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    \(\lim_{x \rightarrow \infty}(\sqrt{x^2-3x}-x)\) \(=\lim_{x \rightarrow \infty}(x\sqrt{1-\frac{3}{x}}-x)\) then expand that \(sqrt\) out via binomial

  15. Jhannybean
    • one year ago
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    When you're simplifying (idk if that's the right word for it) outside of the square root... you have to observe where x is tending to. Since x is moving towards \(\infty\) we can divide out the terms by \(x\) as opposed to \(-x\) if \(x\rightarrow -\infty\) . Does that make sense....

  16. CoconutJJ
    • one year ago
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    Hmm.. I think that makes sense

  17. Jhannybean
    • one year ago
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    Oh wow, typos galore in my latex format in the above post. smh. I hope you understood what I wrote up there LOL

  18. Jhannybean
    • one year ago
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    So basically dividing by \(x\) OUTSIDE the square root \(\equiv\) to dividing by \(x^2\) INSIDE the square root.

  19. CoconutJJ
    • one year ago
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    hmmm but how did you (+x/x) I thought it was (-x)

  20. Jhannybean
    • one year ago
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    What do you mean by (+x/x)?

  21. IrishBoy123
    • one year ago
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    \[x\sqrt{1-\frac{3}{x}}-x = \textstyle x[ 1 + \frac{1}{2}(-\frac{3}{x}) - \frac{1}{8}(\frac{3}{x})^2 +...]-x = ...\] and there's your -3/2

  22. CoconutJJ
    • one year ago
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    |dw:1442968334574:dw|

  23. Jhannybean
    • one year ago
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    Basically if your was \(\lim_{x\rightarrow -\infty}\) then we would have to take the equivalent of x\(^2\) outside the square root, which is \(-x\) /

  24. CoconutJJ
    • one year ago
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    oh so we can use both positive and negative.. I guess..?

  25. Jhannybean
    • one year ago
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    \[\color{red}{\sqrt{x^2} = |x| = \pm x}\] This portion is what you have to pay attention to. This little tid bit lets you know that the simplification of your function everywhere outside the square is dependant on whether your function tends to \(+\infty\) or \(-\infty\)

  26. CoconutJJ
    • one year ago
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    doesn't \[\sqrt{x^2-3x}\] as x-> infinity tend to positive infinity though?

  27. Jhannybean
    • one year ago
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    yes it does, but see how your variables are under the square root sign? that means the variable with the highest power under the square root is x\(^2\), therefore you're going to divide both terms by x\(^2\)

  28. CoconutJJ
    • one year ago
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    hmm.. ok THANK YOU!

  29. Jhannybean
    • one year ago
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    Do you understand it fully?

  30. CoconutJJ
    • one year ago
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    yea.. I guess but I'm still confused on how (-x) turned to (+x/x) shown in the drawing. didn't you divide every term by +x not -x ?

  31. Jhannybean
    • one year ago
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    hint: think about how we came to simplify the original function.

  32. CoconutJJ
    • one year ago
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    |dw:1442969305759:dw|

  33. Jhannybean
    • one year ago
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    original function: \(\color{red}{\sqrt{x^2+3x}-x}\) simplified function: \(\color{blue}{\dfrac{-3x}{\sqrt{x^2-3x}+x}}\)

  34. Jhannybean
    • one year ago
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    somewhere in YOUR simplification process you changed the \(+x\) to a \(-x\)

  35. CoconutJJ
    • one year ago
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    ohhhhhhhhh..... I seeee... sorry.

  36. Jhannybean
    • one year ago
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    No worries, lol

  37. CoconutJJ
    • one year ago
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    okay.. now it makes sense

  38. CoconutJJ
    • one year ago
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    THANKS, once again. :)

  39. Jhannybean
    • one year ago
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    \[\large \lim_{x\rightarrow \infty} \sqrt{x^2-3x}-x\]\[\large \lim_{x\rightarrow \infty} \sqrt{x^2-3x}-x \cdot \frac{\sqrt{x^2-3x}+x}{\sqrt{x^2-3x}+x}\]\[\large \lim_{x\rightarrow \infty} \frac{-3x}{\sqrt{x^2-3x}+x}\]\[\large \lim_{x\rightarrow \infty} \frac{-\dfrac{3x}{x}}{\sqrt{\dfrac{x^2}{x^2} - \dfrac{3x}{x^2}} +\dfrac{x}{x}}\]\[\large \lim_{x\rightarrow \infty}\frac{-3}{\sqrt{1-\dfrac{3}{x}}+1}\]\[=\boxed{-\frac{3}{2}}\]

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