1. **Problem Statement:** Find the limit of the function $$f(x) = \frac{x^2 - 1}{x - 1}$$ as $$x \to 1$$ and determine if the function is continuous at $$x=1$$. 2. **Recall the limit definition:** The limit $$\lim_{x \to a} f(x) = L$$ means as $$x$$ approaches $$a$$ from both sides, $$f(x)$$ approaches $$L$$. 3. **Direct substitution:** Substitute $$x=1$$ into $$f(x)$$: $$f(1) = \frac{1^2 - 1}{1 - 1} = \frac{0}{0}$$ which is indeterminate, so direct substitution fails. 4. **Factor numerator:** Factor $$x^2 - 1$$ as $$(x-1)(x+1)$$: $$f(x) = \frac{(x-1)(x+1)}{x-1}$$ 5. **Simplify expression:** For $$x \neq 1$$, cancel $$(x-1)$$: $$f(x) = x + 1$$ 6. **Evaluate limit:** Now, $$\lim_{x \to 1} f(x) = \lim_{x \to 1} (x + 1) = 1 + 1 = 2$$ 7. **Check continuity:** A function is continuous at $$x=a$$ if: - $$f(a)$$ is defined. - $$\lim_{x \to a} f(x)$$ exists. - $$\lim_{x \to a} f(x) = f(a)$$. Here, $$f(1)$$ is undefined (division by zero), so the function is **not continuous** at $$x=1$$. 8. **Summary:** The limit $$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2$$ exists, but since $$f(1)$$ is undefined, the function has a removable discontinuity at $$x=1$$. **Final answer:** $$\boxed{\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2}$$ and $$f(x)$$ is not continuous at $$x=1$$ because $$f(1)$$ is undefined.