1. **State the problem:** We need to discuss the continuity of the function $$f(x) = \begin{cases}\frac{e^{x} - 1}{e^{x} + 1}, & x \neq 0 \\ 0, & x = 0\end{cases}$$ 2. **Recall the definition of continuity:** A function $f$ is continuous at a point $x=a$ if $$\lim_{x \to a} f(x) = f(a).$$ 3. **Check continuity at $x=0$:** - First, find $f(0)$ which is given as $0$. - Next, find the limit as $x$ approaches $0$: $$\lim_{x \to 0} \frac{e^{x} - 1}{e^{x} + 1}.$$ 4. **Evaluate the limit:** - Use the fact that $e^{0} = 1$. - Substitute $x=0$ directly since numerator and denominator are continuous functions: $$\lim_{x \to 0} \frac{e^{x} - 1}{e^{x} + 1} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0.$$ 5. **Compare limit and function value:** - Since $\lim_{x \to 0} f(x) = 0$ and $f(0) = 0$, the function is continuous at $x=0$. 6. **Check continuity for $x \neq 0$:** - For $x \neq 0$, $f(x)$ is a quotient of continuous functions where the denominator $e^{x} + 1 \neq 0$ for all real $x$. - Therefore, $f(x)$ is continuous for all $x \neq 0$. 7. **Conclusion:** The function $f(x)$ is continuous for all real numbers $x$.