Let $f\colon\mathbb{R}\rightarrow\mathbb{R}$ be a function such that $f(0)=\frac{1}{\pi}$ and $f(x)=\frac{x}{e^{\pi x}-1}$ for $x\ne0$, then

Analysis of Continuity and Differentiability

Function:

\[ f(x) = \begin{cases} \dfrac{x}{e^{\pi x} - 1}, & x \neq 0 \\ \dfrac{1}{\pi}, & x = 0 \end{cases} \]

✅ Continuity at \( x = 0 \):

\[ \lim_{x \to 0} f(x) = \frac{1}{\pi} = f(0) \quad \Rightarrow \quad \text{Function is continuous at } x = 0 \]

✏️ Differentiability at \( x = 0 \):

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = -\frac{1}{2} \]

✅ Final Result:

• Function is continuous at \( x = 0 \)
• Function is differentiable at \( x = 0 \)
• \( f'(0) = \boxed{-\frac{1}{2}} \)