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Math Help - Limits of absolute value function

  1. #1
    DBA
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    Limits of absolute value function

    Hello, I need some help with this

    I need to find the limit of lim x->-2 2-|x| / 2+x

    I need to evaluate the left and right handed limit. For the righthanded limit I plug in

    lim -2+ 2-(x)/2+x,
    now my problem is is to get something that can cancel out, so that I got the limit.

    The lefthanded limit I did as following:
    lim -2- 2-(-x)/2+x = lim -2- 2+x/2+x = 1

    Thanks for any hint.
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  2. #2
    ynj
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    Quote Originally Posted by DBA View Post
    Hello, I need some help with this

    I need to find the limit of lim x->-2 2-|x| / 2+x

    I need to evaluate the left and right handed limit. For the righthanded limit I plug in

    lim -2+ 2-(x)/2+x,
    now my problem is is to get something that can cancel out, so that I got the limit.

    The lefthanded limit I did as following:
    lim -2- 2-(-x)/2+x = lim -2- 2+x/2+x = 1

    Thanks for any hint.
    You don't need to calculate two sides of limit. Since when x->-2, you can assume that x<0.
    so \lim_{x\rightarrow -2}\frac{2-|x|}{2+x}=\lim_{x\rightarrow -2}\frac{2+x}{2+x}=1
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  3. #3
    DBA
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    Sorry, but I do not understand that. How can I asume that x<0 in this case?
    Can you try to explain this in a different way?
    Thanks!
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  4. #4
    ynj
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    Quote Originally Posted by DBA View Post
    Sorry, but I do not understand that. How can I asume that x<0 in this case?
    Can you try to explain this in a different way?
    Thanks!
    Because the limit is only related to the values near -2, that is to say the value between [-2+\delta,+\infty],[-\infty,-2-\delta],\forall\delta>0is totally meaningless. So you can just consider the case when x<0.
    Or you can refer the definition of the limit:
    \forall\epsilon>0,\exists\delta>0,\forall |x-(-2)|<\delta,|f(x)-limit|<\epsilon. In this way, you may understand the "limit"better.
    Since if \delta_0satisfy the definition for some \epsilon>0, then any 0<\delta<\delta_0also satisfy the definition. So the limit is only related to the values near -2.
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  5. #5
    DBA
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    Thanks, I guess I was thinking in the wrong way. It took me a while to understand, but I guess what I did wrong was indeed not to see that in both cases x is smaller than zero and so I get 2+x in bith numerators.

    I am still not sure if I understood the definition of the limit, because I do not know all these characters but I understand the problem to get an answer.
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