1. **State the problem:** Find the value of $k$ such that the function $$f(x) = \begin{cases} \frac{x^2 - 1}{x - 1}, & x \neq 1 \\ k, & x = 1 \end{cases}$$ is continuous at $x = 1$. 2. **Recall the definition of continuity at a point:** A function $f$ is continuous at $x = a$ if $$\lim_{x \to a} f(x) = f(a).$$ 3. **Find the limit as $x \to 1$ of $f(x)$ for $x \neq 1$:** Factor the numerator: $$x^2 - 1 = (x - 1)(x + 1).$$ So, $$f(x) = \frac{(x - 1)(x + 1)}{x - 1}.$$ 4. **Simplify the expression for $x \neq 1$:** Cancel the common factor $x - 1$: $$f(x) = \frac{\cancel{(x - 1)}(x + 1)}{\cancel{x - 1}} = x + 1.$$ 5. **Evaluate the limit:** $$\lim_{x \to 1} f(x) = \lim_{x \to 1} (x + 1) = 1 + 1 = 2.$$ 6. **Set $f(1) = k$ equal to the limit for continuity:** $$k = 2.$$ **Final answer:** $$\boxed{2}$$ This means the function is continuous at $x=1$ if and only if $k=2$.