1. **Problem Statement:** We have a piecewise function $$g(t) = \begin{cases} \frac{1}{2}t + 1 & \text{if } t < 2 \\ \sqrt{t} - 2 & \text{if } t \geq 2 \end{cases}$$ We need to: a. Determine if the limit exists at $t = -2$. b. Determine if the limit exists at $t = 3$. c. Check if $g(t)$ is continuous throughout its domain. --- 2. **Recall the limit definition:** The limit of $g(t)$ as $t$ approaches $a$ exists if and only if the left-hand limit (LHL) and right-hand limit (RHL) at $a$ exist and are equal. 3. **Part a: Limit at $t = -2$** - Since $-2 < 2$, we use the first piece $g(t) = \frac{1}{2}t + 1$ for values near $-2$. - Calculate the limit: $$\lim_{t \to -2} g(t) = \lim_{t \to -2} \left( \frac{1}{2}t + 1 \right) = \frac{1}{2}(-2) + 1 = -1 + 1 = 0$$ - Because the function is defined by the same expression on an open interval around $-2$, the limit exists and equals 0. --- 4. **Part b: Limit at $t = 3$** - Since $3 \geq 2$, we use the second piece $g(t) = \sqrt{t} - 2$ for values near 3. - Calculate the limit: $$\lim_{t \to 3} g(t) = \lim_{t \to 3} (\sqrt{t} - 2) = \sqrt{3} - 2$$ - The function is defined by the same expression on an open interval around 3, so the limit exists and equals $\sqrt{3} - 2$. --- 5. **Part c: Continuity of $g(t)$ throughout its domain** - A function is continuous at a point if: 1. The function is defined at that point. 2. The limit exists at that point. 3. The limit equals the function value at that point. - For $t < 2$ and $t > 2$, each piece is continuous because they are linear and square root functions respectively. - Check continuity at $t = 2$ (the boundary): - Left-hand limit (LHL): $$\lim_{t \to 2^-} g(t) = \lim_{t \to 2^-} \left( \frac{1}{2}t + 1 \right) = \frac{1}{2}(2) + 1 = 1 + 1 = 2$$ - Right-hand limit (RHL): $$\lim_{t \to 2^+} g(t) = \lim_{t \to 2^+} (\sqrt{t} - 2) = \sqrt{2} - 2 \approx 1.414 - 2 = -0.586$$ - Since LHL $\neq$ RHL, the limit at $t=2$ does not exist. - Also, $g(2) = \sqrt{2} - 2 \approx -0.586$. - Because the limit does not exist at $t=2$, $g(t)$ is **not continuous** at $t=2$. --- **Final answers:** - a. The limit at $t = -2$ exists and equals 0. - b. The limit at $t = 3$ exists and equals $\sqrt{3} - 2$. - c. The function $g(t)$ is not continuous at $t = 2$ but continuous elsewhere in its domain.