Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a function such that \(f(0) = \pi\), and
\[
f(x) = \frac{e^{x\pi}}{1 - \pi} \quad \text{for } x \neq 0
\]
Then,
Step-by-step Solution:
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).