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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)=xx2xf(x)=\frac{x}{x^{2}-x}


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The domain of the function is

Df=R{0,1}.\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

limx0xx2x=limx0xx(x1)=limx01x1=1,\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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