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Limit problem using l'hopital's rule

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{x\rightarrow2}\frac{x\sqrt{x-1}-2}{x-2}[/tex]


    2. Relevant equations

    calculate the limit

    3. The attempt at a solution

    When I apply l'hopital's i have to do derivative of the top over derivative of the bottom right?
     
    Last edited: Mar 26, 2012
  2. jcsd
  3. Mar 26, 2012 #2

    SammyS

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    Yes .
     
  4. Mar 26, 2012 #3

    HallsofIvy

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    You could use L'Hopital but I would consider that "overkill". Instead, rationalize the numerator and factor:

    [tex]\frac{x\sqrt{x- 1}- 2}{x- 2}\frac{x\sqrt{x- 1}+ 2}{x\sqrt{x- 1}+ 2}= \frac{x^2(x- 1)- 4}{(x- 2)(x\sqrt{x- 1}+ 2)}= \frac{x^3- x^2- 4}{(x- 1)(x\sqrt{x- 1}+ 2)}[/tex]
    [tex]\frac{(x- 2)(x^2+ x+ 2)}{(x- 2)(x\sqrt{x- 1}- 2)}= \frac{x^2+ x+ 2}{x\sqrt{x- 1}+ 2}[/tex]

    Now, let x= 2.

    (We knew it would factor that way precisely because x= 2 made both numerator and denominator equal to 0.)
     
  5. Mar 26, 2012 #4
    Another alternative:


    [tex]Let \ x \ =\ y^2 + 1.[/tex]



    [tex]As \ \ x\to 2, \ \ y \to 1.[/tex]


    [tex]\lim_{y \to 1} \dfrac{(y^2 + 1)\sqrt{y^2} - 2}{(y^2 + 1) - 2} \ =[/tex]

    [tex]\lim_{y \to 1}\dfrac{|y|(y^2 + 1) - 2}{y^2 - 1} \ = \ [/tex]

    [tex]\lim_{y \to 1}\dfrac{y(y^2 + 1) - 2}{y^2 - 1} \ ** \ = \ [/tex]

    [tex]\lim_{y \to 1}\dfrac{y^3 + y - 2}{y^2 - 1} \ = \ [/tex]

    [tex]\lim_{y \to 1}\dfrac{(y - 1)(y^2 + y + 2)}{(y - 1)(y + 1)} \ = \ [/tex]

    [tex]\lim_{y \to 1}\dfrac{y^2 + y + 2}{y + 1}[/tex]


    Continue . . .





    [tex]** \ Note: \ \ \ As \ \ y \ \ is \ \ positive \ \ near \ \ 1, \ \ |y| = y.[/tex]
     
  6. Mar 26, 2012 #5
    I used L'Hopital's rule, and I think it's actually simpler in this case. The derivative of the denominator is 1, the derivative of the constant 2 is 0 so ...
     
  7. Mar 27, 2012 #6
    ....
     
  8. Mar 28, 2012 #7
    I meant that the derivative of the constant -2 (which appears in the numerator) is 0, so we just have to use the product rule to take the derivative of x*sqrt(x-1). The result,

    √(x-1) + (x/2)*((√(x-1))^(-1))

    ... sorry I don't have practice entering these equations on this system ... but now you can just sub in x = 2 and the limit falls right out.

    √(2-1)+(2/2)*((√(2-1)^(-1))

    √1 + (1)*(1/√1)= 2

    what could be simpler?

    I hope I put in the right number of parens.
     
    Last edited: Mar 28, 2012
  9. Mar 28, 2012 #8
    Yes I solved it guys thanks for the help. By the way the problem stated on solving it using l'hopital's rule
     
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