One–Sided Limits — Question 8

Question 8 - Solution

We are given: $$f(x) = \begin{cases} \frac{x^2 - 4}{x - 2}, & \text{if } x < 2 \\ 3x - 4, & \text{if } x \geq 2 \end{cases}$$

(a) Left-hand limit:

Note: $x^2 - 4 = (x - 2)(x + 2)$, so for $x \neq 2$, $$\frac{x^2 - 4}{x - 2} = x + 2$$ $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 2) = 4$$ $$\boxed{\lim_{x \to 2^-} f(x) = 4}$$

(b) Right-hand limit: $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x - 4) = 6 - 4 = 2$$ $$\boxed{\lim_{x \to 2^+} f(x) = 2}$$

(c) Two-sided limit:

The left-hand and right-hand limits are not equal: $$\lim_{x \to 2^-} f(x) = 4 \neq 2 = \lim_{x \to 2^+} f(x)$$

$$\boxed{\lim_{x \to 2} f(x) \text{ does not exist}}$$

(d) Continuity at $x = 2$:

$\lim_{x \to 2} f(x)$ does not exist , $f(2) = 3(2) - 4 = 6 - 4 = 2$ (defined)

But since the limit does not exist, $f$ is not continuous at $x = 2$.

$$\boxed{f \text{ is not continuous at } x = 2}$$