1. **State the problem:** Analyze the continuity of the piecewise function $$f(x) = \begin{cases}\frac{x^2 - 1}{x - 1}, & x < 1 \\ \sqrt{x+3} - 2, & x \geq 1\end{cases}$$ at $x=1$, examining limits and function values. 2. **Simplify the first piece for $x < 1$: ** Note that $x^2 - 1 = (x-1)(x+1)$. So, $$f(x) = \frac{x^2 - 1}{x - 1} = \frac{(x-1)(x+1)}{x - 1} = x + 1, \quad x \neq 1.$$ 3. **Calculate left-hand limit as $x \to 1^-$: ** Since for $x<1$, $f(x)=x+1$, $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 1) = 1 + 1 = 2.$$ 4. **Calculate right-hand limit as $x \to 1^+$: ** For $x \geq 1$, $f(x) = \sqrt{x + 3} - 2$, $$\lim_{x \to 1^+} f(x) = \sqrt{1 + 3} - 2 = \sqrt{4} - 2 = 2 - 2 = 0.$$ 5. **Evaluate $f(1)$: ** Since $x=1$ uses the second part, $$f(1) = \sqrt{1 + 3} - 2 = 0.$$ 6. **Check continuity at $x=1$: ** - $\lim_{x \to 1^-} f(x) = 2$ - $\lim_{x \to 1^+} f(x) = 0$ - $f(1) = 0$ Since the left and right limits are not equal, the overall limit $\lim_{x \to 1} f(x)$ does not exist, so $f$ is discontinuous at $x=1$. 7. **Continuity from the right: ** Because $\lim_{x \to 1^+} f(x) = f(1) = 0$, $f$ is continuous from the right at $x=1$. 8. **Continuity from the left: ** $\lim_{x \to 1^-} f(x) = 2$ but $f(1)=0$, so $f$ is not continuous from the left at $x=1$. **Final conclusion:** The function $f$ is discontinuous at $x=1$ because the left and right limits differ; it is continuous from the right but not from the left at that point.