Question

f(x)=\frac{x}{x^{2}-x}
removable discontinuities
x = 0
nonremovable discontinuities
x = 1

Explanation:

Step1: Factor denominator

Denominator \( x^2 - x = x(x - 1) \), so \( f(x) = \frac{x}{x(x - 1)} \).

Step2: Find discontinuities

Denominator zero when \( x(x - 1) = 0 \Rightarrow x = 0 \) or \( x = 1 \).

Step3: Check removable discontinuity

Cancel \( x \) ( \( x
eq 0 \) ), \( f(x) = \frac{1}{x - 1} \). Limit at \( x=0 \) exists (-1), so \( x=0 \) is removable.

Step4: Check nonremovable discontinuity

At \( x=1 \), simplified \( f(x) = \frac{1}{0} \), infinite discontinuity. Thus \( x=1 \) is nonremovable.

Answer:

1