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Limit approaching infinity

  1. Sep 17, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]\lim_{\substack{x\rightarrow \infty}}f(x)=\sqrt{3x^2+8x+6}-\sqrt{3x^2+3x+1}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I truly have no idea how to solve this. I know I need to get [tex]x[/tex] in some rational form like [tex]5/x[/tex] but I'm not sure how to do this with the radicals.

    Thanks for any point in the right direction
  2. jcsd
  3. Sep 17, 2007 #2


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    How would you rationalize the fraction if that were in the denominator? Do the same type of thing here.
  4. Sep 17, 2007 #3
    Actually the quotient law of radicals just dawned on me...
    [tex]\frac{\sqrt{a}}{\sqrt{b}} = \sqrt {\frac{a}{b}}[/tex]
  5. Sep 17, 2007 #4


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    Staff Emeritus
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    But you don't have a quotient of radicals! learningphysics suggestion was that you think of this as
    and multiply both numerator and denominator by
    the divide both numerator and denominator by x.
  6. Sep 17, 2007 #5
    Yeah I solved it, I needed that law to solve it this way:

    [tex]\lim_{\substack{x\rightarrow \infty}}f(x)=\sqrt{3x^2+8x+6}-\sqrt{3x^2+3x+1}}[/tex]

    [tex]\lim_{\substack{x\rightarrow \infty}}f(x)=\frac{\sqrt{3x^2+8x+6}-\sqrt{3x^2+3x+1}}{1}[/tex]

    [tex]\lim_{\substack{x\rightarrow \infty}}f(x)=\frac{\sqrt{3x^2+8x+6}-\sqrt{3x^2+3x+1}}{1} \cdot \frac{\sqrt{3x^2+8x+6}+\sqrt{3x^2+3x+1}}{\sqrt{3x^2+8x+6}+\sqrt{3x^2+3x+1}}[/tex]

    [tex]\lim_{\substack{x\rightarrow \infty}}f(x)=\frac{3x^2+8x+6-3x^2-3x-1}{\sqrt{3x^2+8x+6}+\sqrt{3x^2+3x+1}}[/tex]

    [tex]\lim_{\substack{x\rightarrow \infty}}f(x)=\frac{3x^2+8x+6-3x^2-3x-1}{\sqrt{3x^2+8x+6}+\sqrt{3x^2+3x+1}}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}[/tex]

    [tex]\lim_{\substack{x\rightarrow \infty}}f(x)=\frac{5x+5(\frac{1}{x})}{(\sqrt{3x^2+ 8x+6}+\sqrt{3x^2+3x+1})\cdot\sqrt{(\frac{1}{x})^2} }[/tex] <---

    [tex]\lim_{\substack{x\rightarrow \infty}}f(x)=\frac{5+\frac{5}{x}}{\sqrt{3+\frac{8}{x}+\frac{6}{x^2}}+\sqrt{3+\frac{3}{x}+\frac{1}{x^2}}}[/tex]



    [tex]\lim_{\substack{x\rightarrow \infty}}f(x)=\frac{5}{2\sqrt{3}}[/tex]

    Is there an easier way to do it or something?
  7. Sep 17, 2007 #6
    Rationalizing that you get:

    [tex]\lim_{\substack{x\rightarrow \infty}}f(x)=\frac{5\sqrt{3}}{6}[/tex]
  8. Sep 17, 2007 #7


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    Easier? Not really: this is pretty much the best "tool" for dealing with indeterminate differences. Some people call this method "multiplication by a conjugate factor". The idea is to exploit "the difference of two squares" to eliminate the radicals and (one hopes) to simplify the expression. It can also be used on limits of rational functions with differences in the denominator.
  9. Oct 11, 2007 #8
    Jwill, can you explain to me why you multiplied the numerator and denominator by 1/x in the 5th step please?
  10. Oct 11, 2007 #9
    It's basically a clever from of one to solve the limit. You just find the highest (nth degree) and multiply the numerator and denominator by 1/x^n.
  11. Oct 12, 2007 #10

    Gib Z

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    The work only seems complex because you showed unnecessary working in my opinion. You would only need lines 3, 5 and 7 to justify your answer.
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