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Math Help - Just want to verify my proof of a limit

  1. #1
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    Just want to verify my proof of a limit

     \lim_{x\rightarrow 2} \frac{x^2+x-6}{x-2}= 5

    Proof: given any   \epsilon >0 choose  \delta =\epsilon

    Ignore the *******< br/ > *******

    then if  \left | x-a \right | < \delta
     \Rightarrow  \left | x-2 \right |<\epsilon \Rightarrow  \left | x+3-5 \right | <\epsilon  <br />
 \Rightarrow \left | \frac{ (x+3)(x-2)}{x-2} -5 \right | <\epsilon   \Rightarrow  \left | \frac{x^2+x-6}{x-2}-5\right | < \epsilon  \Rightarrow  \left | f(x)-L \right | < \epsilon

    Therefore  \lim_{x\rightarrow 2} \frac{x^2+x-6}{x-2}= 5
    I want to make sure if I'm right thanks guys! This is just the proof no scratch work
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  2. #2
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    Re: Just want to verify my proof of a limit

    Yes, that's a perfectly good proof. The basic idea, of course, is that, for x not equal to 2, \frac{x^2+ x- 6}{x- 2}= \frac{(x+ 3)(x- 2)}{x- 2}= x+ 3 so this problem is really "linear" and you can do it that way.
    Thanks from mathisfun26
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  3. #3
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    Re: Just want to verify my proof of a limit

    Quote Originally Posted by mathisfun26 View Post
     \lim_{x\rightarrow 2} \frac{x^2+x-6}{x-2}= 5
    Proof: given any   \epsilon >0 choose  \delta =\epsilon   then if  \left | x-a \right | < \delta

    Actually I would do this differently form reply #2.

    Note that \text{If }x\ne 2\text{ then }\left|\frac{x^2+x-6}{x-2}-5\right|=|x-2|.

    So you are correct we can use \delta=\varepsilon.
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  4. #4
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    Re: Just want to verify my proof of a limit

    Plato's reply points out a certain flaw in the original proof:

    what if |x - 2| = 0?

    your proof should start with:

    if  0 < |x - 2| < \delta....

    this allows you to multiply and divide by (x - 2) without fear of it being undefined.
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