1. **State the problem:** We are given a piecewise function: $$f(x) = \begin{cases}\frac{x^2 - x}{x^2 - 1} & \text{if } x \neq 1 \\ 1 & \text{if } x = 1\end{cases}$$ We want to analyze the function at $a = 1$. 2. **Simplify the expression for $x \neq 1$:** Factor numerator and denominator: $$x^2 - x = x(x - 1)$$ $$x^2 - 1 = (x - 1)(x + 1)$$ So, $$f(x) = \frac{x(x - 1)}{(x - 1)(x + 1)}$$ 3. **Cancel common factors:** Since $x \neq 1$, we can cancel $(x - 1)$: $$f(x) = \frac{x}{x + 1}$$ 4. **Evaluate the limit as $x$ approaches 1:** $$\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x}{x + 1} = \frac{1}{1 + 1} = \frac{1}{2}$$ 5. **Check the value of the function at $x = 1$:** Given $f(1) = 1$. 6. **Conclusion:** The limit of $f(x)$ as $x$ approaches 1 is $\frac{1}{2}$, but $f(1) = 1$. Therefore, $f$ is not continuous at $x = 1$ because the limit and the function value differ. **Final answer:** $$\lim_{x \to 1} f(x) = \frac{1}{2} \neq f(1) = 1$$ Hence, $f$ is discontinuous at $x=1$.