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.
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.
$\displaystyle \begin{align*}f(x) & = \dfrac{\sqrt{(5x^2+1)(x^2-4x+4)}}{x^2-2x} \\ & = \dfrac{\sqrt{(5x^2+1)(x-2)^2}}{x^2-2x} \\ & = \dfrac{|x-2|\sqrt{5x^2+1}}{x(x-2)} \\ & = \begin{cases}-\dfrac{\sqrt{5x^2+1}}{x} & x<2 \\ \dfrac{\sqrt{5x^2+1}}{x} & x>2\end{cases}\end{align*}$
Notes:
$\displaystyle f(2)$ does not exist.
$\displaystyle \displaystyle \lim_{x \to 2^-} f(x) = -\dfrac{\sqrt{21}}{2}$
$\displaystyle \displaystyle \lim_{x \to 2^+} f(x) = \dfrac{\sqrt{21}}{2}$
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.
When you have $\displaystyle u^2 = v$, then solving for $\displaystyle u$, you get $\displaystyle u = \pm \sqrt{v}$. This is because the square root symbol always produces the nonnegative root. So, when you take $\displaystyle \sqrt{u^2}$, you get only the nonnegative root: $\displaystyle |u|$.
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.
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} $