Question

Multiple Choice Which of the following points of discontinuity of $$f(x)=\frac{x(x-1)(x-2)^{2}(x+1)^{2}(x-3)^{2}}{x(x-1)(x-2)(x+1)^{2}(x-3)^{3}}$$ is not removable? \(\begin{array}{ll}{(\mathbf{A}) x=-1} & {(\mathbf{B}) x=0} \\ {(\mathbf{D}) x=2} & {(\mathbf{E}) x=3}\end{array} \quad(\mathbf{C}) x=1\)

Short answer

The option (E) \(x=3\) is the point of discontinuity of the function that is not removable.

Step-by-step solution

Simplify the Function

First, simplify the function by cancelling the common factors in the numerator and the denominator: $$f(x)=\frac{x(x-1)(x-2)^{2}(x+1)^{2}(x-3)^{2}}{x(x-1)(x-2)(x+1)^{2}(x-3)^{3}} = \frac{(x-2)}{(x-3)}$$ Now, function \(f(x)\) is undefined for \(x=3\).

Find if the discontinuity is removable

A point \(x=a\) is a removable discontinuity if the limit \(\lim_{x\to a} f(x)\) exists and is finite. We need to find the limit of \(f(x)\) as \(x\) approaches 3. But it can be seen that as \(x\) approaches 3 the function \(f(x)\) tends to infinity. So, the point \(x=3\) is a non-removable discontinuity.