Answer:
(a) See below.
(b) x = 0 or x = 1
(c) x = 0 removable, x = 1 non-removable
Step-by-step explanation:
Given rational function:
[tex]f(x)=\dfrac{\ln |x-1|}{x}[/tex]
Part (a)
Substitute x = 2 into the given rational function:
[tex]\begin{aligned}\implies f(2) & =\dfrac{\ln |2-1|}{2}\\\\ & =\dfrac{\ln 1}{2}\\\\ & =\dfrac{0}{2}\\\\ & =0\end{aligned}[/tex]
Therefore, as the function is defined at x = 2, the function is continuous at x = 2.
Part (b)
Given interval: [-2, 2]
Logs of negative numbers or zero are undefined. As the numerator is the natural log of an absolute value, the numerator is undefined when:
|x - 1| = 0 ⇒ x = 1.
A rational function is undefined when the denominator is equal to zero, so the function f(x) is undefined when x = 0.
So the function is discontinuous at x = 0 or x = 1 on the interval [-2, 2].
Part (c)
x = 1 is a vertical asymptote. As the function exists on both sides of this vertical asymptote, it is an infinite discontinuity. Since the function doesn't approach a particular finite value, the limit does not exist. Therefore, x = 1 is a non-removable discontinuity.
A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal.
[tex]\implies f(0)=\dfrac{\ln |0-1|}{0}=\dfrac{\ln 1}{0}=\dfrac{0}{0}[/tex]
Therefore, there is a hole at x = 0.
The removable discontinuity of a function occurs at a point where the graph of a function has a hole in it. Therefore, x = 0 is a removable discontinuity.
