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Limit at inifinity

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Can you guys just check to see if I'm right?

    [tex] f(x) = \frac{2x-\sqrt{4x^2-5x+300}}{1} [/tex]

    3. The attempt at a solution


    [tex] \frac{2x - \sqrt{4x^2-5x+300}}{1} * \frac{2x+ \sqrt{4x^2-5x+300}}{2x+ \sqrt{4x^2-5x+300}}} [/tex]

    [tex] \frac{4x^2 - 4x^2 + 5x - 300}{2x+ \sqrt{4x^2-5x+300}} [/tex]

    [tex] \frac{\frac{5x-300}{x}}{\frac{2x+ \sqrt{4x^2-5x+300}}{x}} [/tex]

    [tex] \frac{5- \frac{300}{x}}{2 + \sqrt{ \frac{4x^2 - 5x + 300}{x^2}}} [/tex]

    [tex] \frac{5 + 0}{2 + \sqrt{ \frac{4x^2}{x^2} - \frac{5x}{x^2} + \frac{300}{x^2}}} [/tex]

    [tex] \frac{5}{2 + \sqrt{4 - 0 +0}} [/tex]

    [tex] \frac{5}{4} [/tex]
     
    Last edited: Oct 22, 2009
  2. jcsd
  3. Oct 21, 2009 #2
    yes, it's 5/4.. generally, the technique is to divide out by the term with the highest degree
     
  4. Oct 21, 2009 #3

    tiny-tim

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    Hi Zhalfirin88! :smile:

    It's \lim_{x\rightarrow \infty} … [tex] \lim_{x\rightarrow \infty}[/tex] :wink:
     
  5. Oct 21, 2009 #4
    Oh, thanks tiny-tim :)
     
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