One–Sided Limits — Question 5

Question 5 - Solution

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

(a) Left-hand limit:

As $x \to 3^-$, we are approaching 3 from values less than 3. The expression is: $$f(x) = \frac{1}{x - 3}$$

As $x \to 3^-$, $x - 3 \to 0^- \Rightarrow \frac{1}{x - 3} \to -\infty$

$$\boxed{\lim_{x \to 3^-} f(x) = -\infty}$$

(b) Right-hand limit:

As $x \to 3^+$, the function becomes: $$f(x) = \sqrt{x - 3}$$

As $x \to 3^+$, $x - 3 \to 0^+ \Rightarrow \sqrt{x - 3} \to 0$

$$\boxed{\lim_{x \to 3^+} f(x) = 0}$$

(c) Does the limit exist?

Since $$\lim_{x \to 3^-} f(x) = -\infty \quad \text{and} \quad \lim_{x \to 3^+} f(x) = 0$$ the two one-sided limits are not equal (and one diverges), so the limit $\lim_{x \to 3} f(x)$ does not exist.

$$\boxed{\text{Limit does not exist at } x = 3}$$