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Homework Help: Help with a limit?

  1. Jun 4, 2004 #1
    I’m having trouble understanding this limit

    Lim x->0 (x^2-2|x|)/x

    I think its undefined but lost marks for that answer on my class assignment.

    Also, Why can’t I simplify to x-2?
  2. jcsd
  3. Jun 4, 2004 #2


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    The limit exists if and only if the left limit and right limit exist, and are equal to each other.

    [tex]\lim _{x \rightarrow {0}^{+}} \frac{x^2 - 2|x|}{x}[/tex]

    [tex]= \lim _{x \rightarrow {0}^{+}} \frac{x^2 - 2x}{x}[/tex]

    [tex]= \lim _{x \rightarrow {0}^{+}} x - 2[/tex]

    [tex]= -2[/tex]

    [tex]\lim _{x \rightarrow {0}^{-}} \frac{x^2 - 2|x|}{x}[/tex]

    [tex]= \lim _{x \rightarrow {0}^{-}} \frac{x^2 + 2x}{x}[/tex]

    [tex]= \lim _{x \rightarrow {0}^{-}} x + 2[/tex]

    [tex]= 2[/tex]

    Therefore, the limit does not exist.
    Last edited: Jun 4, 2004
  4. Jun 4, 2004 #3
    You can also see it graphically by plotting the function. There is a large discontinuity (gap) at x=0.
  5. Jun 4, 2004 #4
    Say x = -x. Then
    [tex]\frac{(-x)^2 - 2|-x|}{-x} = \frac{x^2 - 2x}{-x} = 2 - x[/tex]​
    As you can see, you can't JUST simplify to x - 2. If x > 0, then you can simplify to x - 2.

    What happens when x approaches 0 from the positive direction? What happens when x approaches 0 from the negative direction? You should see (if you haven't already) that the limit is undefined.
  6. Jun 4, 2004 #5
    because |x|/x is not equal to 1. It is equal to 1 if x >0 and -1 if x < 0. You can simplify that to x - 2|x|/x. That would help in finding the limit. In order to find the limit you will have to break that up into a piecewise function and than look at the limit as it approaches 0 from the left and the right. And ill give you a hint, there is no limit as x --> 0.
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