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Math Help - Graph Sketching of Limits

  1. #1
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    Graph Sketching of Limits

    Sketch a possible graph of the following:

    f(-2) exsists, lim x --> -2+ f(x)= f(-2), but lim x --> -2 f(x) does not exist.

    I'm new to Latex. I'm working on it.



    I came with the crude graph below, the series leaning toward the right comes from the right and the series from the left comes from the left.. What stumps me is the meaning of lim x--> -2+ f(x)= f(-2).
    I don't understand, or I don't think I do, what it means.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post
    Sketch a possible graph of the following:

    f(-2) exsists, lim x --> -2+ f(x)= f(-2), but lim x --> -2 f(x) does not exist.

    I'm new to Latex. I'm working on it.



    I came with the crude graph below, the series leaning toward the right comes from the right and the series from the left comes from the left.. What stumps me is the meaning of lim x--> -2+ f(x)= f(-2).
    I don't understand, or I don't think I do, what it means.
    \lim_{x \to -2^+}f(x) this means the limit as x approaches -2 from the right. this is called the right hand limit.

    for the limit to exist, the left and right hand limits must be equal, so just make sure as the graph approaches -2 from the left it is not equal to the value of the graph when it approaches from the right. make sure when we are approaching from the right, we are defined at x = -2.

    you graph does the job, but it is weird. if you extended it, and only left the dots at the ends, it would be better. how about this graph? the dotted line is not a part of the graph, i merely put it there to show you that the points were in line
    Attached Thumbnails Attached Thumbnails Graph Sketching of Limits-discontinuous.jpg  
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  3. #3
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    I don't know how I knew you would answer Jhevon, but I did somehow. lol

    I see what you're saying.

    The part that stumped me simply meant that the graph ended at -2.
    Okay.

    The graph was just to get a faint idea.

    Thank yoU!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post
    I don't know how I knew you would answer Jhevon, but I did somehow. lol
    well, i just hate to disappoint

    I see what you're saying.

    The part that stumped me simply meant that the graph ended at -2.
    Okay.

    The graph was just to get a faint idea.

    Thank yoU!
    you're welcome. just notice how i had one circle shaded and one unshaded. i did that on purpose
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