One–Sided Limits — Question 7

Question 7 - Solution

We are given: $$f(x) = \begin{cases} 3x + 1, & \text{if } x < 2 \\ x^2 - 4, & \text{if } x \geq 2 \end{cases}$$

(a) Left-hand limit as $x \to 2^-$:

Use the definition for $x < 2$: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (3x + 1) = 3(2) + 1 = 7$$ $$\boxed{\lim_{x \to 2^-} f(x) = 7}$$

(b) Right-hand limit as $x \to 2^+$:

Use the definition for $x \geq 2$: $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 - 4) = 2^2 - 4 = 0$$ $$\boxed{\lim_{x \to 2^+} f(x) = 0}$$

(c) Two-sided limit:

$$\lim_{x \to 2^-} f(x) = 7, \quad \lim_{x \to 2^+} f(x) = 0$$ Since the one-sided limits are not equal: $$\boxed{\lim_{x \to 2} f(x) \text{ does not exist}}$$

(d) Value of the function at 2:

Since $x = 2 \geq 2$, use the second case: $$f(2) = 2^2 - 4 = 0$$ $$\boxed{f(2) = 0}$$