Continuity — Question 3

Question 3 - Solution

We are given: $$f(x) = \begin{cases} \dfrac{x^2 - 4}{x - 2}, & x \neq 2 \\ k, & x = 2 \end{cases}$$

To make $f(x)$ continuous at $x = 2$, the following condition must hold: $$\lim_{x \to 2} f(x) = f(2)$$

Step 1: Compute the limit as $x \to 2$

Factor the numerator: $$x^2 - 4 = (x - 2)(x + 2)$$

For $x \neq 2$, the function simplifies to: $$f(x) = x + 2$$

So, $$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 2) = 4$$

Step 2: Match the function value

Continuity requires: $$f(2) = k = 4$$

Final Answer: $$\boxed{k = 4}$$