# Chapter 1 - Limits and Their Properties - 1.4 Exercises: 44

$f(x)$ has two vertical asymptotes (non-removable discontinuity) at $x=2$ and $x=-2.$

#### Work Step by Step

$f(x)=\dfrac{x}{x^2-4}=\dfrac{x}{(x-2)(x+2)}\to x\ne2$ and $x\ne-2$ Hence since the denominator is zero at both values, the function has non-removable discontinuities at $x=2$ and $x=-2.$

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