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Math Help - continuous function

  1. #1
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    continuous function

    k(x) is a piecewise function

    if k(x)= (x^2-2b+b^2)/(x-b) if x does not equal a
    = 5 if x=a

    then which of the following is true about k(x):
    1) limx->a k(x) exists
    2) k(x) is continuous at x=a
    3) k(a) exists

    so far i think that k(a) will exist because it will equal 5 right? realli not sure about the other two.
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  2. #2
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    Hi chukie! First, let me make your notation easier to read:

    Quote Originally Posted by chukie View Post
    k(x) is a piecewise function

    k(x) = <br />
\left\{\begin{array}{rl}<br />
\frac{x^2-2b+b^2}{x-b}, & \text{if }x\neq a\\<br />
                                5, & \text{if }x=a<br />
\end{array}\right.

    then which of the following is true about k(x):

    1) \lim_{x\to a}k(x)\text{ exists}

    2) k(x) \text{ is continuous at }x=a

    3) k(a) \text{ exists}

    so far i think that k(a) will exist because it will equal 5 right? realli not sure about the other two.
    You are correct that k is defined at a.

    For the others:

    1. When taking a limit, we only care about what happens to a function as it approaches a certain value, not what happens at that value. So, the limit of k as x\to a will be the same as the limit of the first "piece."

    2. For a function f(x) to be continuous at a point x = c, three things must be true:

    * f(c)\text{ is {d}efined}

    * \lim_{x\to c}f(x) \text{ exists}

    * \lim_{x\to c}f(x) = f(c)

    You should know the first 2 by answering the other two parts of this problem. So, take the limit of k(x) as x\to a. Then, if the value of this limit is k(a) = 5, then k is continuous at a, and it will be discontinuous otherwise.

    You should be able to find which value for b makes the function continuous.
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  3. #3
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    thanks so much! ur awesome
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