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Math Help - Limits/Piecewise-defined function

  1. #1
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    Limits/Piecewise-defined function

    Okay, so I'm new here, so I'm not quite sure what the rules are for introductions and whatnot. This isn't an introduction, just...I don't know, how you introduce the problem that you need help on. I've just started Calculus and we have a quiz soon, yet I don't understand the current homework assignment. Also, I'm not quite sure how to make all of the correct letters so, uhm, please, bear with me for a moment.

    Instructions:

    a) Draw the graph of f.
    b) Determine limx-->c+ f(x) and limx-->c- f(x)
    c) Does limx-->c f(x) exist? If so, what is it? If not, explain.


    Question:

    1. c=1, f(x)={ 1 - x^2, x does not equal -1; 2, x=-1}

    Problem: Now, I did the previous problems with almost no complications. However, they all had a greater than or less than, etc, sign on them. The equal signs here are what throws me off...


    Instructions:

    a) Draw the graph of f.
    b) At what points c in the domain of f does limx-->c f(x) exist?
    c) At what points c does only the left-hand limit exist?
    d) At what points c does only the right-hand limit exist?

    Problems:

    2. f(x)={ sinx, -2π ≤ x < 0; cosx, 0 ≤ x ≤ π}

    3. f(x)={cosx, -π ≤ x < 0; secx, 0 ≤ x ≤ π }

    I'm not quite sure, but I think it was the sin and cos that threw me off...the ones with regular numbers don't really seem so bad. I'm sure if I could get some help on these, I'll be able to figure out the others with no problem.


    Uhm...yeah, thank you in advance for any help. (Oh, and π <-- those things, are pi. It was really the best I could do...)
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  2. #2
    MHF Contributor red_dog's Avatar
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    To find if f has limit at the point c you have to verify if \displaystyle\lim_{x\nearrow c}f(x)=\lim_{x\searrow c}f(x).
    Here are the skethces for 1) and 2)
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  3. #3
    MHF Contributor red_dog's Avatar
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  4. #4
    MHF Contributor red_dog's Avatar
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    For 3) the function is correct? Is \sec x or \sin x?
    Because \sec x does not exists for \displaystyle x=\frac{\pi}{2}
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