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Math Help - Limit help ? Limit Continuous ? part 1

  1. #1
    Newbie mathfun9's Avatar
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    Smile Limit help ? Limit Continuous ? part 1

    Can someone solve for this problem ? Thank you so much !!!


    1. Define h(2) so that h(x)=( X^2 + 3X - 10 )/( X-2 ) IS CONTINUOUS AT X =1


    2. wHAT VALUE SHOULD BE ASSIGNED to b to make the function

    g (x) = { x^3 , x < 1/2
    { bx^2 ,x > and equal 1/2



    3. Explain why the limits exist and give their values . Support with graphing utility !!

     lim      [ ( 1 + cos X ) / 2 ]
    X--> 0


    Thank you !!!
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  2. #2
    Eater of Worlds
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    #2:

    g(1/2)=\frac{1}{8}

    Find b so that \lim_{x\to \frac{1}{2}^{+}}bx^{2}=\frac{1}{8}
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  3. #3
    MHF Contributor

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    Can you factor x^2  + 3x  10?
    If so, then if you do not understand at that point, you are really over you head!
    Please do not drown in your own lack of understanding.
    \frac{{x^2  + 3x - 10}}{{x - 2}} = \frac{{\left( {x - 2} \right)\left( {x + 5} \right)}}{{x - 2}} = x + 5\,,\,x \ne 2.
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  4. #4
    Newbie mathfun9's Avatar
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    yes ...!!

    yes I understand ..that part ..I just need to know how to we define ..the continuous at x=2 !!!

    If I 've said anything wrong ..I'm sorry about that !!!
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  5. #5
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    Quote Originally Posted by mathfun9 View Post
    yes I understand ..that part ..I just need to know how to we define ..the continuous at x=2 !!!

    If I 've said anything wrong ..I'm sorry about that !!!
    Find the limit as x approaches 2. Let's name it L. Then define a piece-wise function, such that:

    \left\{ \begin{array}{rcl} \frac{x^2+3x-10}{x-2} & x\neq2 \\<br />
L & x = 2 \end{array} \right.
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