Letf(x) = x^2−25 / x-5. Notice thatf(5) is not defined because of the denominator.How should we definef(5) so thatfis continuous at 5?

Answer 1

Factor the numerator. Cancel the factor (x-5) from the numerator and denominator to get f(x) = x + 5 (for x ≠ 5). Then, plug in x = 5 to get the y-value 10.

Let f(5) = 10 and the function is continuous.

Answer 2

What are the requirements for continuity?

1. The limit exists
2. The function in question has a defined value at that point
3. The values from 1) and 2) are equal to each other.

Here, the limit as x approaches 5 is 10. To make this continuous, then, we need to make f(5) = 10.