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# For the following exercise, determine the point(s), if any, at which the function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other. $f(x)=\frac{x}{x^{2}-x}$

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The domain of the function is

$\mathcal{D}_f = \mathbb{R} \setminus \{ 0, 1 \}.$

The function is a rational function, thus continuous everywhere on its domain. Since 0 and 1 are not in the domain, the function is discontinuous at these points. Lets classify these discontinuities

• Since

$\lim_{x \to 0} \dfrac{x}{x^2-x} = \lim_{x \to 0} \dfrac{x}{x(x-1)} = \lim_{x \to 0} \dfrac{1}{x-1} = -1,$

we see that the function has a $\text{\color{#4257b2} removable discontinuity at 0. \color{default} • Since $$\lim_{x \to 1^\pm} \dfrac{x}{x^2-x} = \lim_{x \to 1^\pm} \dfrac{1}{x-1} = \dfrac{1}{\text{ } 0^\pm} = \pm \infty,$$ we conclude that the function has an \color{#4257b2} infinite discontinuity at 1. }$

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