Given: \[ f(x) = \begin{cases} \pi, & x = 0 \\ \dfrac{e^{x\pi}}{1 - \pi}, & x \neq 0 \end{cases} \] We are to analyze continuity and differentiability at \( x = 0 \).
Step 1: Check Continuity at \( x = 0 \)
A function is continuous at \( x = 0 \) if: \[ \lim_{x \to 0} f(x) = f(0) \] We know:
\( f(0) = \pi \)
For \( x \neq 0 \), \( f(x) = \dfrac{e^{x\pi}}{1 - \pi} \)
So, \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \dfrac{e^{x\pi}}{1 - \pi} = \dfrac{e^{0}}{1 - \pi} = \dfrac{1}{1 - \pi} \] But \( f(0) = \pi \)
So, \[ \lim_{x \to 0} f(x) = \dfrac{1}{1 - \pi} \neq \pi = f(0) \] \( f(x) \) is not continuous at \( x = 0 \)
So, not differentiable either (since differentiability implies continuity).