One–Sided Limits — Question 10

Question 10 - Solution

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

(a) Left-hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 - 3x + 2) = 1^2 - 3(1) + 2 = 0$$ $$\boxed{\lim_{x \to 1^-} f(x) = 0}$$

(b) Right-hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (\sqrt{x + 3} - 2) = \sqrt{4} - 2 = 2 - 2 = 0$$ $$\boxed{\lim_{x \to 1^+} f(x) = 0}$$

(c) Two-sided limit:

Since both one-sided limits exist and are equal: $$\boxed{\lim_{x \to 1} f(x) = 0}$$

(d) Continuity at $x = 1$:

We already have: $$\lim_{x \to 1} f(x) = 0$$

Now evaluate the function at $x = 1$: $$f(1) = \sqrt{1 + 3} - 2 = \sqrt{4} - 2 = 2 - 2 = 0$$

Since: $f(1)$ is defined , $\lim_{x \to 1} f(x) = f(1)$

We conclude: $$\boxed{f \text{ is continuous at } x = 1}$$