One–Sided Limits — Question 6

Question 6 - Solution

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

(a) Left-hand limit:

For $x < 0$, use: $$\lim_{x \to 0^-} \frac{\sin(\pi x)}{x}$$

Using the identity $\lim_{x \to 0} \frac{\sin(\pi x)}{x} = \pi$ (from standard limit $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$): $$\lim_{x \to 0^-} \frac{\sin(\pi x)}{x} = \pi$$ $$\boxed{\lim_{x \to 0^-} f(x) = \pi}$$

(b) Right-hand limit:

For $x \geq 0$, use: $$\lim_{x \to 0^+} \frac{1}{x + 1} = \frac{1}{1} = 1$$ $$\boxed{\lim_{x \to 0^+} f(x) = 1}$$

(c) Two-sided limit:

Since: $$\lim_{x \to 0^-} f(x) = \pi \quad \text{and} \quad \lim_{x \to 0^+} f(x) = 1$$

These are not equal, so the limit does not exist: $$\boxed{\lim_{x \to 0} f(x) \text{ does not exist}}$$