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Thread: limits

  1. #1
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    limits

    lim x→2g(x)
    whereg(x) = ( x − 2 /x^ 2 − 4) if x 2
    g(x)= 1 if x = 2
    this confuses me any help would be nice ....
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  2. #2
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    Re: limits

    As it is written, your function is $\displaystyle \begin{align*} x - \frac{2}{x^2} - 4 \end{align*}$, but I suspect you meant $\displaystyle \begin{align*} \frac{x - 2}{x^2 - 4} \end{align*}$. Which one is it? If it's the latter, you MUST use brackets where they are needed!
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    Re: limits

    yes it's the latter
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    Re: limits

    hmmm the part that confuses me is I break it down to (x-2)/(x-2)(x+2) => 1/(x+2)

    then sub in x = 2 to become 1/4

    what confuses me is the part if x ≠ 2 and g(x)= 1 if x = 2

    I'm having trouble with that concept
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  5. #5
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    Re: limits

    Quote Originally Posted by bee77 View Post
    hmmm the part that confuses me is I break it down to (x-2)/(x-2)(x+2) => 1/(x+2)
    then sub in x = 2 to become 1/4
    what confuses me is the part if x ≠ 2 and g(x)= 1 if x = 2
    I'm having trouble with that concept
    NEVER, NEVER, NEVER Say that. It will earn you a failing mark.

    The notation $x\approx a$ means that $x$ is close to $a$.
    So $\displaystyle \lim_{x\to a}f(x)=L$ means that if $x\approx a$ but $x\ne a$ then $f(x)\approx L$.

    Now $\dfrac{x-2}{x^2-4}=\dfrac{1}{x+2}$ unless $x=2$.

    So $\displaystyle \lim_{x\to 2}\dfrac{x-2}{x^2-4}=\displaystyle \lim_{x\to 2}\dfrac{1}{x+2}=\dfrac{1}{4}$ because by definition of limit $x\ne 2$ so $\dfrac{x-2}{x^2-4}=\dfrac{1}{x+2}$
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  6. #6
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    Re: limits

    $\dfrac{x-2}{x^2-4}$ is not defined at $x=2$ because you are dividing by zero (which is not a valid operation). However, they defined $g(x) = \begin{cases}\dfrac{x-2}{x^2-4} & x\neq 2 \\ 1 & x=2\end{cases}$ to mean that if $x\neq 2$, then $g(x) = \dfrac{x-2}{x^2-4}$ and $g(2) = 1$.

    A limit of a function has nothing to do with the value of that function at that point. This is what is known as a removable discontinuity.
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    Re: limits

    Thanks Plato ,I'll remember that sub part for my upcoming exam ...and the explanation of the concepts with the close to symbol really helps ,
    cheers!!
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    Re: limits

    Thanks SlipEternal ,so the g(2)=1 would simply imply a different part of the graph ...to the x-2/x^2-4 = 1/4 limit...?
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  9. #9
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    Re: limits

    I don't know what you mean by "imply a different part of the graph". The graph of y= g(x) is the graph of 1/(x+ 2) except that the point (1, 1/4) is removed and the point (1, 1) is added. But you take x closer and closer to 1, but not equal to 1, (x, y) on the graph gets closer and closer to (1, 1/4).
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    Re: limits

    Quote Originally Posted by Prove It View Post
    [bee77], you MUST use brackets where they are needed!
    Quote Originally Posted by bee77 View Post
    hmmm the part that confuses me is I break it down to (x-2)/(x-2)(x+2) => 1/(x+2)
    (x - 2)/[(x - 2)(x + 2)] => 1/(x + 2),  x \ne 2

    Quote Originally Posted by bee77 View Post
    x-2/x^2-4 = ...
    (x - 2)/(x^2 - 4) =

    bee77, do you see how the brackets are needed?
    Last edited by greg1313; Jul 17th 2017 at 07:06 AM.
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  11. #11
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    Re: limits

    $g(x)=\left\{\begin{matrix}
    \dfrac{x-2}{x^2-4}\, & \, x\ne2\\
    1 \, & \, x=2
    \end{matrix}\right.$

    graph of $g(x)$ attached ... can you now visualize why \lim_{x \to 2} g(x) = \dfrac{1}{4}?
    Attached Thumbnails Attached Thumbnails limits-piecewisef.jpg  
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  12. #12
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    Re: limits

    That makes it better in my head on the graph skeeter ,
    Thanks !
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    Re: limits

    Thanks HallsofIvy , I was just overthinking it in my head , thanks for the help
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