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Thread: Can someone please explain : Vertical Asymptotes

  1. #1
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    Can someone please explain : Vertical Asymptotes

    For the function f(x) = sqrt((5x^2+1)(x^2-4x+4)) / (x^2-2x)

    Why is x =0 the only vertical asymptote? The denominator becomes 0 when x = 0 , x =2.
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  2. #2
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    Re: Can someone please explain : Vertical Asymptotes

    Because it is a removable discontinuity at $x=2$, not a vertical asymptote.

    There is a factor of $x-2$ in the numerator as well as the denominator. So, while the function cannot take the value $x=2$, there is no value of x near 2 where the function approaches positive or negative infinity.
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    Re: Can someone please explain : Vertical Asymptotes

    Quote Originally Posted by SlipEternal View Post
    Because it is a removable discontinuity at $x=2$, not a vertical asymptote.

    There is a factor of $x-2$ in the numerator as well as the denominator. So, while the function cannot take the value $x=2$, there is no value of x near 2 where the function approaches positive or negative infinity.
    Oops! I meant it is a jump discontinuity, not a removable one.
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    Re: Can someone please explain : Vertical Asymptotes

    Thanks I think I know Why, but I can't seem to simplify it so the factors cancel!!

    Is it jump discontinuity because if I take the limit as x approches 2, they go to different values?
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    Re: Can someone please explain : Vertical Asymptotes

    Quote Originally Posted by lc99 View Post
    Thanks I think I know Why, but I can't seem to simplify it so the factors cancel!!

    Is it jump discontinuity because if I take the limit as x approches 2, they go to different values?
    \begin{align*}f(x) & = \dfrac{\sqrt{(5x^2+1)(x^2-4x+4)}}{x^2-2x} \\ & = \dfrac{\sqrt{(5x^2+1)(x-2)^2}}{x^2-2x} \\ & = \dfrac{|x-2|\sqrt{5x^2+1}}{x(x-2)} \\ & = \begin{cases}-\dfrac{\sqrt{5x^2+1}}{x} & x<2 \\ \dfrac{\sqrt{5x^2+1}}{x} & x>2\end{cases}\end{align*}

    Notes:
    f(2) does not exist.
    \displaystyle \lim_{x \to 2^-} f(x) = -\dfrac{\sqrt{21}}{2}
    \displaystyle \lim_{x \to 2^+} f(x) = \dfrac{\sqrt{21}}{2}
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    Re: Can someone please explain : Vertical Asymptotes

    Thanks, i've always been confused with how you can just create the piece-wise function.

    How do i know when i can do that? Is it done when the function in the form abs(x) /x? Can it be x / abs(x)?

    And, does the (x-2) comes out of the radical function as absolute value because it's squared inside the radical function?

    Thanks for your help Slip!
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    Re: Can someone please explain : Vertical Asymptotes

    Quote Originally Posted by lc99 View Post
    Thanks, i've always been confused with how you can just create the piece-wise function.

    How do i know when i can do that? Is it done when the function in the form abs(x) /x? Can it be x / abs(x)?

    And, does the (x-2) comes out of the radical function as absolute value because it's squared inside the radical function?

    Thanks for your help Slip!
    You create a piecewise function when it is appropriate to the problem. There may be all sorts of reasons. But when you have a function with equal left and right limits at a point where the function has no existence as a mathematical object, a piecewise function (or substitution by a different function) may make sense.
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    Re: Can someone please explain : Vertical Asymptotes

    Quote Originally Posted by lc99 View Post
    Thanks, i've always been confused with how you can just create the piece-wise function.

    How do i know when i can do that? Is it done when the function in the form abs(x) /x? Can it be x / abs(x)?

    And, does the (x-2) comes out of the radical function as absolute value because it's squared inside the radical function?

    Thanks for your help Slip!
    When you have u^2 = v, then solving for u, you get u = \pm \sqrt{v}. This is because the square root symbol always produces the nonnegative root. So, when you take \sqrt{u^2}, you get only the nonnegative root: |u|.
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    Re: Can someone please explain : Vertical Asymptotes

    In this example, since the x-2 cancels, does this mean that there is a limit that exists but the value of x=2 doesn't exist. So we have to do piece wise? What does x-2 cancelling in this problem indicate?
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    Re: Can someone please explain : Vertical Asymptotes

    Quote Originally Posted by lc99 View Post
    In this example, since the x-2 cancels, does this mean that there is a limit that exists but the value of x=2 doesn't exist. So we have to do piece wise? What does x-2 cancelling in this problem indicate?
    Go back to basics.
    More often then not, poorly trained mathematics instructors tell students that
    if $f(x)=\dfrac{x^3-8}{x-2}=x^2+2x+4$ but of course that is not true. It is true if s/he says when $x\ne 2$.

    Take the function $h(x)=\begin{cases}\dfrac{x^3-8}{x-2} &: x\ne 2 \\ 16 &: x= 2\end{cases}$

    We hope that you can see that the function $h$ is not continuous at $x=2$ because
    $\displaystyle{{\lim _{x \to {2^ - }}}h(x) = {\lim _{x \to {2^ + }}}h(x) }= 12$ BUT $h(2)=16$

    So we have a discontinuity at $x=2$ suppose we redefined the finction:
    $h(x)=\begin{cases}\dfrac{x^3-8}{x-2} &: x\ne 2 \\ 12 &: x= 2\end{cases}$ now the function is continuous at $x=2$.

    Do you see that the discontinuity can and has been removed by redefining the function at $x=2~?~?.$
    Thus we call this a removable discontinuity.
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  11. #11
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    Re: Can someone please explain : Vertical Asymptotes

    Quote Originally Posted by lc99 View Post
    In this example, since the x-2 cancels, does this mean that there is a limit that exists but the value of x=2 doesn't exist. So we have to do piece wise? What does x-2 cancelling in this problem indicate?
    No. As I demonstrated, the limit from the left does not equal the limit from the right. $\dfrac{|x|}{x}\neq 1$. So the $x-2$ does not "cancel" directly. You need to consider when x is positive and when it is negative and how the absolute value affects the sign of the result. You find:

    $\dfrac{|x|}{x} = \begin{cases}-1 & x <0\\1 & x>0\end{cases} $
    Last edited by SlipEternal; Nov 14th 2017 at 02:44 AM.
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