Continuity — Question 9

Question 9 - Solution

We want $f$ to be continuous at $x = 2$, so we must have: $$\lim_{x \to 2} f(x) = f(2) = k$$

First, compute the limit: $$f(x) = \frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}$$

For $x \neq 2$, we can cancel the common factor: $$f(x) = x + 2$$

Therefore: $$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 2) = 4$$

To make $f$ continuous at $x = 2$, we set: $$\boxed{k = 4}$$