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I have a question about the value of limit.

  1. Oct 5, 2009 #1
    I evaluated a limit which is [limit x approaches 2 (-2x^2+4x/x-2)]
    I plugged value 2 into the limit. So, -2(4)+8/2-2 and I got 0/0.
    However, I came up to three possible final answer and I am not sure which is the real final answer.
    I have thought about it, and I asked for people's advice, but they all had different answers. So here is the three answer which came out from the discussion with people, 1, 2, 3.
    "The answer is A)Does not exist" 1 said.
    "No, the answer is B)0" this is No.2 and my idea.
    "Hey, how about C)1?" 3 insisted.
    So I came here to hear what people think about this. I will tell you what I tried. About answer A from person number 1, as you know, if I got an answer 7/0, the limit would be DNE(does not exists). So why not 0/0 can't be DNE?
    Then how about 0? 2 and I think this is the answer but....
    Is it 1 from person number 3, because 1/1 is 1, 2/2 is 1,3/3 is 1 and so forth?
    Please help.
     
  2. jcsd
  3. Oct 5, 2009 #2
    You need to factor the numerator. You will see that a factor of the numerator is the factor that is in the denominator. These will cancel and you will be able to just plug in the 2 and get a real answer.
     
  4. Oct 5, 2009 #3
    When an expression evaluates to something of the form 0/0, it is called an indeterminate quotient. There are more advanced methods (usually seen midway through Calculus I) that can handle these better. It is unfortunately the case that nothing about the limit can be deduced from the knowledge that something an indeterminate quotient of the form 0/0. The limit may exist (i.e. converges to a limit L), be infinitely divergent, or be just plain undefined. Note that converging to a limit and diverging to an infinity are both defined limits while only converging is a limit that exists. Stating that a limit does not exist is not fully answering the question.

    All limits below are indeterminate (0/0) and as can be seen each is different in behavior.

    [tex]\lim_{x \rightarrow 0} \frac{x^2}{x^2} = 1.[/tex]

    [tex]\lim_{x \rightarrow 0} \frac{x}{x^3}= \infty.[/tex]

    [tex]\lim_{x \rightarrow 0} \frac{x^3-x^2}{x} = 0.[/tex]

    [tex]\lim_{x \rightarrow 2} \frac{x^2-4}{x^2+x-6} = \frac{4}{5}.[/tex]

    --Elucidus
     
  5. Oct 6, 2009 #4

    HallsofIvy

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    None of those answers is correct. The fact that the numerator is 0 when x= 2 tells you that x- 2 is a factor of the denominator and so can be cancelled. Like w3390 said, factor the numerator and cancel.
     
  6. Oct 6, 2009 #5
    Now I see what was wrong with my calculation!
    (-2x)(x-2)/(x-2) and by deletling (x-2), (-2x) left. So the answer is -4.
    I thank all for good advices! Every answer was very helpful.
     
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