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Math Help - Limits of piecewise function

  1. #1
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    Limits of piecewise function

    Lim where f (x) = 2 (x+1) if x <3
    x->2 4 if x=3
    x^2 - 1 if x> 3


    please tell me you have to write out what the left and right limits are to find out if they are equal?



    left limit is 6 and right is 3 so the answer is DNE right?!
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    Re: Limits of piecewise function

    If \lim_{x \to c^-} f(x) \ne \lim_{x \to c^+} f(x) , then \lim_{x \to c} f(x) does not exist.
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    Re: Limits of piecewise function

    Quote Originally Posted by habibixox View Post
    Lim where f (x) = 2 (x+1) if x <3
    x->2 4 if x=3
    x^2 - 1 if x> 3


    please tell me you have to write out what the left and right limits are to find out if they are equal?



    left limit is 6 and right is 3 so the answer is DNE right?!
    Is this what you are trying to evaluate?

    \displaystyle \lim_{x \to 2} f(x) where \displaystyle f(x) = \begin{cases}2(x + 1)\textrm{ if }x < 3 \\ 4 \textrm{ if }x = 3 \\ x^2 - 1 \textrm{ if }x > 3\end{cases}
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    Re: Limits of piecewise function

    If, in fact, you are trying to find the limit at 2 then you can ignore everything that is not close to 2. In particular, f(x)= 2(x+ 1) for x "anywhere near 2" (close to 2 is certainly less than 3). You do NOT use " x^2- 1" to evaluate since that is true only for x> 3.
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    Re: Limits of piecewise function

    yes
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  6. #6
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    Re: Limits of piecewise function

    why wouldnt you test out the limits?
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    Re: Limits of piecewise function

    would i be completely wrong if i showed what i did at first on a quiz?

    i showed how the right limit doesn't = the left limit so i wrote in conclusion the limit doesn't exist because 3 is not equal to 6
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    Re: Limits of piecewise function

    \displaystyle f(x) = \begin{cases}2(x + 1)\textrm{ if }x < 3 \\ 4 \textrm{ if }x = 3 \\ x^2 - 1 \textrm{ if }x > 3\end{cases}

    note that 2 < 3 ...

    \displaystyle \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 6
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    Re: Limits of piecewise function

    Quote Originally Posted by habibixox View Post
    would i be completely wrong if i showed what i did at first on a quiz?

    i showed how the right limit doesn't = the left limit so i wrote in conclusion the limit doesn't exist because 3 is not equal to 6
    Well, you are not completely wrong to show it now, but you certainly were wrong to do that on the test.

    Since the limit is at x= 2, not any where near 3,
    \lim_{x\to 2^-} f(x)= \lim_{x\to 2^-} 2(x+1)= 2(3)= 6
    \lim_{x\to 2^+} f(x)= \lim_{x\to 2^+} 2(x+1)= 2(3)= 6
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