1. Given the piecewise function: $$f(x) = \begin{cases} 2 & \text{if } x=0 \\ \frac{x}{\sin(\pi x)} & \text{if } x \neq 0 \end{cases}$$ 2. To check continuity at $x=0$, we need: $$\lim_{x \to 0} f(x) = f(0)$$ 3. Calculate the limit: $$\lim_{x \to 0} \frac{x}{\sin(\pi x)}$$ 4. Use the substitution $u = \pi x$, so as $x \to 0$, $u \to 0$: $$\lim_{u \to 0} \frac{\frac{u}{\pi}}{\sin u} = \frac{1}{\pi} \lim_{u \to 0} \frac{u}{\sin u}$$ 5. Recall that $\lim_{u \to 0} \frac{\sin u}{u} = 1$, so: $$\lim_{u \to 0} \frac{u}{\sin u} = 1$$ 6. Therefore: $$\lim_{x \to 0} \frac{x}{\sin(\pi x)} = \frac{1}{\pi}$$ 7. For continuity at $x=0$, set: $$f(0) = 2 = \lim_{x \to 0} f(x) = \frac{1}{\pi}$$ 8. Since $2 \neq \frac{1}{\pi}$, the function is not continuous at $x=0$ unless $f(0)$ is redefined to $\frac{1}{\pi}$. Final values: - $p = 2$ - $d = 0$ - $II = \frac{1}{\pi}$ (limit value) - $f(x) = \begin{cases} 2 & x=0 \\ \frac{x}{\sin(\pi x)} & x \neq 0 \end{cases}$ Hence, the function is continuous at $x=0$ if $f(0) = \frac{1}{\pi}$.