Question

Which of the following function has a removable discontinuity at the given point?

a) f(x)=x/|x| at a=0

b) f(x)=(x²+x-6)/(x³-3x²+2) at a=0

c) f(x)=(x²+x-6)/(x³-3x²+2) at a=2

d) f(x)=x/(x-2) at a=2

e) f(x)=x/(x-2) at a=0

Answer 1

1. $\lim_{x\to0^+}\frac{x}{|x|}=1, \lim_{x\to0^-}\frac{x}{|x|}=-1$ by definition of |x|, thus this function does not have a removable discontinuity.
2. $\lim_{x\to0}\frac{x^2+x-6}{x^3-3x^2+2}=\frac{-6}{2}=-3$, the limit exists and thus the discontinuity is removable at x=0.
3. $\lim_{x\to 2}\frac{x^2+x-6}{x^3-3x^2+2}=\lim_{x\to 2}\frac{(x-2)(x+3)}{(x-1)(x^2-2x-2)}=0$, as the expression in the denominator has a non-zero limit at x=2 (the limit is $2^3-3\cdot4+2=-2$), so the discontinuity is removable.
4. $\lim_{x\to 2} \frac{x}{x-2} = \pm\infty$, as the expression in denominator approaches zero and the expression in numerator approaches 2. The discontinuity is not removable.
5. $\lim_{x\to 0} \frac{x}{x-2} = 0$, as the expression in the denominator approaches a non zero value (it approaches -2). Thus the discontinuity is removable.