1. State the problem: We have a piecewise function: $$f(x) = \begin{cases} x^2 + 1 & \text{if } x < 1 \\ (x - 2)^2 & \text{if } x \geq 1 \end{cases}$$ Find the left-hand limit $\lim_{x \to 1^-} f(x)$, the right-hand limit $\lim_{x \to 1^+} f(x)$, and determine whether the limit $\lim_{x \to 1} f(x)$ exists. 2. Find the left-hand limit $\lim_{x \to 1^-} f(x)$: For $x < 1$, $f(x) = x^2 + 1$. Evaluate at $x=1$: $$\lim_{x \to 1^-} f(x) = 1^2 + 1 = 1 + 1 = 2$$ 3. Find the right-hand limit $\lim_{x \to 1^+} f(x)$: For $x \geq 1$, $f(x) = (x - 2)^2$. Evaluate at $x=1$: $$\lim_{x \to 1^+} f(x) = (1 - 2)^2 = (-1)^2 = 1$$ 4. Determine if $\lim_{x \to 1} f(x)$ exists: Since $\lim_{x \to 1^-} f(x) = 2$ and $\lim_{x \to 1^+} f(x) = 1$, the left-hand and right-hand limits are not equal. Therefore, the limit $\lim_{x \to 1} f(x)$ does not exist. Final answers: - $\lim_{x \to 1^-} f(x) = 2$ - $\lim_{x \to 1^+} f(x) = 1$ - $\lim_{x \to 1} f(x)$ does not exist.