One–Sided Limits — Question 9

Question 9 - Solution

We are given: $$f(x) = \begin{cases} \frac{\sin(x)}{x}, & \text{if } x < 0 \\ 2x + 1, & \text{if } x \geq 0 \end{cases}$$

(a) Left-hand limit as $x \to 0^-$:

Since $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$, and this applies as $x \to 0^-$ as well: $$\boxed{\lim_{x \to 0^-} f(x) = 1}$$

(b) Right-hand limit as $x \to 0^+$:

$$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2x + 1) = 1$$ $$\boxed{\lim_{x \to 0^+} f(x) = 1}$$

(c) Two-sided limit:

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

(d) Value at the point:

Use the definition for $x \geq 0$: $$f(0) = 2(0) + 1 = 1$$ $$\boxed{f(0) = 1}$$