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find the value of x at which f is not continuous and whether each such value is a removable discontinuity: (a) f(x) = abs(x) / x (b) f(x) = (x^2 + 3x) / (x + 3) (c) f(x) = (x - 2) / (abs(x) - 2)
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in each of these cases, f will fail to be continuous when the denominator is zero, so for (b), f is discontinuous at x=-3. The discontinuity is removable if you can factor things in a way that gets rid of the discontinuity. For example, (b) factors into \[\frac{x(x+3)}{(x+3)} = x\], so the discontinuity is removable.
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