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Thread: continuity

  1. #1
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    continuity

    a is a real number.
    f:R--->R, f(x)= 2x+a,x>=3
    16-ax^2, x<3

    If this function is continious find lim f(x)
    x-->a?

    I found a=1
    6+a=16-9a
    a=1
    when x--->1
    16-x^2=15
    and found the limit as 15.But book says 16-a.
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  2. #2
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    Re: continuity

    The answer is 15 as you determined.

    Whoever wrote the answer key wasn't paying attention.
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  3. #3
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    Re: continuity

    Many Thanks.
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  4. #4
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    Re: continuity

    Quote Originally Posted by kastamonu View Post
    a is a real number.
    f:R--->R, f(x)= 2x+a,x>=3
    16-ax^2, x<3
    If this function is continious find lim f(x) .
    $f(x)=\begin{cases}2x+a & x\ge 3 \\16-ax^2 & x<3\end{cases}$____$\begin{cases}\displaystyle{\lim _{x \to {3^ + }}}f(x) = 6 + a & \\\displaystyle{\lim _{x \to {3^ - }}}f(x) = 16-9a & \end{cases}$

    a=?
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  5. #5
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    Re: continuity

    a=1
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  6. #6
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    Re: continuity

    Quote Originally Posted by kastamonu View Post
    a is a real number.
    f:R--->R, f(x)= 2x+a,x>=3
    16-ax^2, x<3

    If this function is continious find lim f(x)
    x-->a?


    I found a=1

    6+a=16-9a
    a=1
    when x--->1
    16-x^2=15
    and found the limit as 15.But book says 16-a.
    Since a= 1, 16- a is 15
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