Question

In Exercises \(19-24,(\) a) find each point of discontinuity. (b) Which of the discontinuities are removable? not removable? Give reasons for your answers. $$f(x)=\left\\{\begin{array}{ll}{\frac{1}{x-1},} & {x<1} \\ {x^{3}-2 x+5,} & {x \geq 1}\end{array}\right.$$

Short answer

The function \(f(x)\) has a non-removable discontinuity at \(x = 1\)

Step-by-step solution

- Find the limits as x approaches 1 from left and right

Before checking if the function is continuous at \(x=1\), we need to find the limit of \(f(x)\) as \(x\) approaches 1 from the left (denoted as \(x \to 1^-\)), and as \(x\) approaches 1 from the right (denoted as \(x \to 1^+\)). For \(x \to 1^-\), use the part of the function defined for \(x < 1\), which is \(\frac{1}{x-1}\). For \(x \to 1^+\), use the part of the function defined for \(x \geq 1\), which is \(x^3 - 2x + 5\). Using limit rules, it's found that \(\lim_{x \to 1^-}f(x)\) is undefined due to division by zero in \(\frac{1}{x-1}\), but \(\lim_{x \to 1^+}f(x)\) equals 4.

- Identify and classify the point of discontinuity

For a function to be continuous at a point, the limit of the function must exist at that point (and equal the function's value there). It's clear that since \(\lim_{x \to 1^-}f(x)\) doesn't exist while \(\lim_{x \to 1^+}f(x)\) does, \(f(x)\) has a point of discontinuity at \(x = 1\). This type of discontinuity, where the function is undefined at the point of discontinuity, is called a non-removable discontinuity. So at \(x = 1\), \(f(x)\) has a non-removable discontinuity.