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Finding limit at infinity

  1. Oct 9, 2013 #1
    1. Evaluate lim x[itex]\rightarrow[/itex]-[itex]\infty[/itex] [itex]\sqrt{x^2+x+1}[/itex]+x.The answer is -[itex]\frac{1}{2}[/itex].

    2. Relevant equations

    None.

    3. The attempt at a solution

    I multiplied by the conjugate first, so it turns into

    lim x[itex]\rightarrow[/itex]-[itex]\infty[/itex] [itex]\frac{(x^2+x+1)-x^2}{\sqrt{x^2+x+1}-x}[/itex]


    = lim x[itex]\rightarrow[/itex]-[itex]\infty[/itex] [itex]\frac{x+1}{\sqrt{x^2+x+1}-x}[/itex]

    I divide by 1/x on the top, and 1/√x2 on the bottom.

    lim x[itex]\rightarrow[/itex]-[itex]\infty[/itex] [itex]\frac{\frac{x}{x}+\frac{1}{x}}{\sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}+ \frac{1}{x^2} }-\frac{x}{x}}[/itex]

    = lim x[itex]\rightarrow[/itex]-[itex]\infty[/itex] [itex]\frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}
    +\frac{1}{x^2}} -1}[/itex]


    At this point, this is all the algebra I can do. So now I have to plug in the -[itex]\infty[/itex].

    When x goes to -[itex]\infty[/itex] into [itex]\frac{1}{x}[/itex], I get 0. Same with [itex]\frac{1}{x^2}[/itex], I also get 0.

    So wouldn't that make my equation

    [itex]\frac{1+0}{(√1+0+0)-1}[/itex]? My answer would be undefined then, not -[itex]\frac{1}{2}[/itex]...
     
  2. jcsd
  3. Oct 9, 2013 #2

    Dick

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    x is NEGATIVE. If that's the case then sqrt(f)/x=(-sqrt(f/x^2)). Think about examples with numbers.
     
  4. Oct 9, 2013 #3
    Yes, but how does that apply to my equation?
     
  5. Oct 9, 2013 #4
    Even if x is negative, [itex]\frac{1}{x}[/itex] would be so small that it would be insignificant.
     
  6. Oct 9, 2013 #5

    Dick

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    You are screwing up a sign with x being negative. You should be getting 1/(-1-1). The sign on the square root is wrong.
     
  7. Oct 9, 2013 #6

    Dick

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    No idea what you are talking about. (-1)*sqrt(2) is -sqrt(2*(-1)^2). It's not sqrt(2*(-1)^2).
     
  8. Oct 9, 2013 #7
    I think my algebra is right. What step did I do wrong?
     
  9. Oct 9, 2013 #8

    Dick

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    Your algebra is right if x>0. Your algebra is dead wrong if x<0. I've told you.
     
  10. Oct 9, 2013 #9

    Office_Shredder

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    When you divide by 1/x on the top, and [itex] 1/\sqrt{x^2}[/itex] on the bottom you have multiplied your expression by -1.
     
  11. Oct 9, 2013 #10

    Dick

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    And you also converted ##1/\sqrt{x^2}## into 1/x when you multiplied the second term. They just plain aren't equal.
     
  12. Oct 9, 2013 #11
    Okay, thanks I got it.
     
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