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Simple limit (verification)

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data
    1z6uzp1.png


    2. Relevant equations

    N/A (L'Hospital (maybe)??

    3. The attempt at a solution


    so this is what i tried,
    209hzc1.jpg

    since i got infinity - infininty, i tried to convert to a fraction then went on from there to get 2

    but using wolfram it gave me -2, mind that it used l'hoapital after converting to a fraction, i didn't find it necessary to apply L'H, since i simplified and didn't get an intermidate form.


    wolfram:: http://www.wolframalpha.com/input/?i=lim%20x-%3E%20-infinity%20%28sqrt%28%28x^2%29%2B2x%29-%28sqrt%28%28x^2%29-2x%29&t=wvg01
     

    Attached Files:

  2. jcsd
  3. Sep 19, 2010 #2
    Your last step is not really justified.

    [tex] \sqrt{x^{2} +2x}+ \sqrt{x^{2}-2x }= |x| \left(\sqrt{1 + \frac{2}{x}} + \sqrt{1 -\frac{2}{x}} \right) [/tex]

    Since x gets more and more negative

    |x| = -x ( in this case)
     
  4. Sep 19, 2010 #3
    abs(x)abs so i can use it as -x and x and see if they boht approch the same #?
     
  5. Sep 19, 2010 #4

    In your specific case they will not approach the same number as you can see. That is why wolfram got -2.

    |x| = -x if x <0
    |x| = x if x >= 0


    In you case x was going to negative infinity. So it only make sense to put abs(x) = -x.
    In your solution you simiply put [tex] |x| = x[/tex] which is not justified since you are bringing stuff out of the square root.
     
  6. Sep 21, 2010 #5
    hey.. so how can i work with the absolute's? help...
     
  7. Sep 21, 2010 #6
    What do you mean ?
     
  8. Sep 21, 2010 #7
    [tex]
    \sqrt{x^{2} +2x}+ \sqrt{x^{2}-2x }= |x| \left(\sqrt{1 + \frac{2}{x}} + \sqrt{1 -\frac{2}{x}} \right)
    [/tex]


    from here can i find the limit like i take in to account like split into -x and x and find the values... i did find them but they are -2 and 2... where im i going wrong


    EDIT: you said something about using -x...why is that...
     
  9. Sep 22, 2010 #8

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I don't see how that will do any good at all since you still have that "|x|" going to infinity and the difference going to 0.

    What I recommend, even though you don't really have a fraction, is to "rationalize the numerator":
    [tex]\sqrt{x^2+ 2x}- \sqrt{x^2- 2x}= \left(\sqrt{x^2+ 2x}- \sqrt{x^2- 2x}\right)\frac{\sqrt{x^2+ 2x}+ \sqrt{x^2- 2x}}{\sqrt{x^2+ 2x}+ \sqrt{x^2- 2x}}[/tex]
    [tex]= \frac{x^2+ 2x- (x^2- 2x)}{\sqrt{x^2+ 2x}+ \sqrt{x^2- 2x}}[/tex]
    [tex]= \frac{4x}{\sqrt{x^2+ 2x}+ \sqrt{x^2- 2x}}[/tex]

    Now divide both numerator and denominator by x, making it [itex]x^2[/itex] inside the square roots:
    [tex]\frac{4}{\sqrt{1+ 2/x}+ \sqrt{1- 2/x}}[/tex]

    Now take the limit as x goes to negative infinity.
     
  10. Sep 22, 2010 #9
    Okay I will explain in more details.

    [tex]\sqrt{x^2+ 2x}- \sqrt{x^2- 2x}= \left(\sqrt{x^2+ 2x}- \sqrt{x^2- 2x}\right)\frac{\sqrt{x^2+ 2x}+ \sqrt{x^2- 2x}}{\sqrt{x^2+ 2x}+ \sqrt{x^2- 2x}}[/tex]
    [tex]= \frac{x^2+ 2x- (x^2- 2x)}{\sqrt{x^2+ 2x}+ \sqrt{x^2- 2x}}[/tex]
    [tex]= \frac{4x}{\sqrt{x^2+ 2x}+ \sqrt{x^2- 2x}}[/tex]

    This is all good. You did that part excellently.


    [tex] \frac{4x}{ \left( \sqrt{ x^{2} \left(1+ \frac{2}{x}\right)}+ \sqrt{ x^{2} \left(1- \frac{2}{x} \right)} \right)}[/tex]

    Take the [tex]x^2[/tex] term out
    [tex]=\frac{4x}{ \left|x \right| \left( \sqrt{1+ 2/x}+ \sqrt{1- 2/x} \right)}[/tex]

    The absolute value sign is because...
    [tex] \sqrt{x^2} = \left|x \right| [/tex]

    We know that |x| is defined as follows

    |x| = -x if x<0
    |x| = x if x >= 0

    In your limit x <0 since you are making x approach negative infinity . x would be negative so -x would make x positive. Do you understand this ?

    And that is why you should replace |x| with -x

    That is ...
    [tex]=\frac{4x}{-x \left( \sqrt{1+ 2/x}+ \sqrt{1- 2/x} \right)}[/tex]
    [tex]=\frac{-4}{ \sqrt{1+ 2/x}+ \sqrt{1- 2/x}}[/tex]

    And now you can happily take your limit to and you will see that it approaches [tex] \frac{-4}{2} = -2[/tex]






    That is exactly what OP did which was incorrect.

    Following your solution leads to a limit going to [tex] \frac{4}{2} =2[/tex] which is not correct!
     
  11. Sep 22, 2010 #10
    ╔(σ_σ)╝ thank you so much i see where i went wrong...(I always make these mistake)
     
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