At what values of x is f(x) = (x^2 - 1)/(x^2 - 3x + 2) discontinuous? Classify each discontinuity as removable or non-removable and justify the answer.

Step-by-step solution

1. 1

   Locate the suspects: the zeros of the denominator. A rational function is continuous everywhere its denominator is nonzero, so the only possible trouble spots come from

   $x^2-3x+2=(x-1)(x-2)=0\ \Longrightarrow\ x=1,\ x=2.$

2. 2

   Factor the numerator too - that is what decides the classification. Since $x^2-1=(x-1)(x+1)$,

   $f(x)=\frac{(x-1)(x+1)}{(x-1)(x-2)}.$

   A discontinuity is removable exactly when the offending factor cancels, because then the two-sided limit exists and only the function value is missing.

3. 3

   Handle $x=1$: the factor cancels, so the limit exists. For $x\neq 1$ the function equals $\dfrac{x+1}{x-2}$, hence

   $\lim_{x\to 1}f(x)=\frac{1+1}{1-2}=-2,$

   while $f(1)$ is undefined. The limit exists but does not equal the (missing) value, so $x=1$ is a removable discontinuity - a hole at $(1,-2)$. Redefining $f(1)=-2$ would patch it.

4. 4

   Handle $x=2$: the factor survives. Near $x=2$ the reduced form is $\dfrac{x+1}{x-2}$ with numerator approaching $3\neq 0$ and denominator approaching $0$, so

   $\lim_{x\to 2^-}f(x)=-\infty,\qquad \lim_{x\to 2^+}f(x)=+\infty.$

   Numerically, $f(1.999999)\approx -3\times 10^{6}$ and $f(2.000001)\approx 3\times 10^{6}$. No finite limit exists, so no redefinition can fix it: $x=2$ is non-removable (an infinite discontinuity with a vertical asymptote).

5. 5

   State the conclusion with justification. $f$ is discontinuous only at $x=1$ and $x=2$; $x=1$ is removable because the common factor $x-1$ cancels and the limit $-2$ exists, and $x=2$ is non-removable because the one-sided limits are infinite and of opposite sign.