# Thread: Finding y value of point of discontinuity

1. ## Finding y value of point of discontinuity

The directions for the problem state: The following rational functions each have a hole at the given x value. Find the corresponding y value of this point of discontinuity.

I can see how the x value given makes the numerator and denominator zero. But I am confused because in past problems we have done, the holes always had to be a cancelled out factor after we simplified the problem. How do I go about finding y in a problem like this?

Thanks for any help

2. ## Re: Finding y value of point of discontinuity rationalize the denominator

$\dfrac{x-5}{\sqrt{2x-1}-3} \cdot \dfrac{\sqrt{2x-1}+3}{\sqrt{2x-1}+3} = \dfrac{(x-5)(\sqrt{2x-1}+3)}{(2x-1)-9} = \dfrac{\cancel{(x-5)}(\sqrt{2x-1}+3)}{2\cancel{(x-5)}}= \dfrac{\sqrt{2x-1}+3}{2}$

now determine the limit of the last expression as $x \to 5$

Thank you!

4. ## Re: Finding y value of point of discontinuity

Please note that this is NOT a valid procedure if x = 5. (x-5)/(x-5) is NOT 1 when x = 5.

5. ## Re: Finding y value of point of discontinuity Originally Posted by TKHunny Please note that this is NOT a valid procedure if x = 5. (x-5)/(x-5) is NOT 1 when x = 5.
But please also note that if $x\to 5$ means that $\large x\ne 5$, that is $x$ is close to $5$ but not ever there.
In the formal definition we say $(\exists \delta>0)[0<|x-5|<\delta \Rightarrow$
Please realize that the set $0<|x-5|<\delta$ is $\{x|x \in(5-\delta,5)\cup(5,5+\delta)\}$ so $x\ne 5$.

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