Is the function f(x) = (x^2 - 1)/(x - 1) continuous at x = 1?

Is the function f(x) = (x^2 - 1)/(x - 1) continuous at x = 1?

• A. Yes
• B. No
• C. It is continuous at all x ≠ 1
• D. It is undefined at x = 1 and not continuous there

Answer: (B)

Continuity at a specific point requires the function to be defined at that point and the limit as you approach the point to match the function value there. Here, the function is not defined at x = 1 because the denominator is zero, so continuity at that exact point cannot hold. If you simplify the expression for x ≠ 1, you get f(x) = x + 1, so the limit as x approaches 1 is 2. Since f(1) does not exist, the function cannot be continuous at x = 1, even though the limit exists. This situation shows a removable discontinuity (a hole) at x = 1.

Continuity at a specific point requires the function to be defined at that point and the limit as you approach the point to match the function value there. Here, the function is not defined at x = 1 because the denominator is zero, so continuity at that exact point cannot hold. If you simplify the expression for x ≠ 1, you get f(x) = x + 1, so the limit as x approaches 1 is 2. Since f(1) does not exist, the function cannot be continuous at x = 1, even though the limit exists. This situation shows a removable discontinuity (a hole) at x = 1.