One-Sided Limits — Question 1

Question 1

Let the function $f$ be defined as:

$$f(x) = \begin{cases} \dfrac{x^2 - 4}{x - 2}, & \text{if } x < 2 \\ x + 2, & \text{if } x \geq 2 \end{cases}$$

Evaluate: $$\lim_{x \to 2^-} f(x) \quad \text{and} \quad \lim_{x \to 2^+} f(x)$$

State whether $\lim_{x \to 2} f(x)$ exists.

Question 1 - Solution

We analyze the left-hand and right-hand limits separately.

Left-hand limit:

For $x < 2$, $$f(x) = \frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}$$

For $x \neq 2$, the $x - 2$ cancels: $$f(x) = x + 2$$

So: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 2) = 4$$

Right-hand limit:

For $x \geq 2$, we use the second piece: $$f(x) = x + 2 \Rightarrow \lim_{x \to 2^+} f(x) = 4$$

Conclusion:

Since both one-sided limits exist and are equal: $$\lim_{x \to 2} f(x) = \boxed{4}$$