1. **State the problem:** We are given two functions $f(x)$ and $g(x)$ from graphs and asked to find various limits involving these functions. 2. **Recall limit rules:** The limit of a sum is the sum of the limits if both limits exist. 3. **Analyze each limit:** (a) $\lim_{x \to 2} [f(x) + g(x)]$: - From the graphs, $f(2)$ approaches 3 (since the line continues upward through (1,1) and beyond, assume $f(2)=3$). - $g(2)$ approaches 1 (assuming linear continuation). - So, $\lim_{x \to 2} f(x) = 3$, $\lim_{x \to 2} g(x) = 1$. - Therefore, $\lim_{x \to 2} [f(x) + g(x)] = 3 + 1 = 4$. (b) $\lim_{x \to 0} [f(x) + g(x)]$: - $f(x)$ has a jump at 0: left limit $f(0^-) = 1$ (open circle at (0,1)), right limit $f(0^+) = 1$ (line passes through (1,1)). So $\lim_{x \to 0} f(x) = 1$. - $g(x)$ is continuous at 0 with $g(0) = 1$. - So, $\lim_{x \to 0} [f(x) + g(x)] = 1 + 1 = 2$. (c) $\lim_{x \to 0^+} [f(x) + g(x)]$: - $f(0^+) = 1$, $g(0^+) = 1$. - Sum is $1 + 1 = 2$. (d) $\lim_{x \to 0^-} [f(x) + g(x)]$: - $f(0^-) = 1$, $g(0^-) = 1$. - Sum is $1 + 1 = 2$. (e) $\lim_{x \to 2} \frac{f(x)}{3 + g(x)}$: - $f(2) = 3$, $g(2) = 1$. - Denominator $3 + 1 = 4$. - Limit is $\frac{3}{4}$. (f) $\lim_{x \to 2} \frac{1 + g(x)}{f(x)}$: - Numerator $1 + 1 = 2$. - Denominator $f(2) = 3$. - Limit is $\frac{2}{3}$. (g) $\lim_{x \to 0^+} \sqrt{f(x)}$: - $f(0^+) = 1$. - Limit is $\sqrt{1} = 1$. (h) $\lim_{x \to 0^-} \sqrt{f(x)}$: - $f(0^-) = 1$. - Limit is $\sqrt{1} = 1$. (i) $\lim_{x \to -2} f(x)$: - Left side limit $f(-2^-)$ is the open circle at ( -2, 1 ), so limit from left is 1. - Right side limit $f(-2^+)$ is the closed dot at ( -2, -2 ), so limit from right is -2. - Since left and right limits differ, $\lim_{x \to -2} f(x)$ does not exist. **Final answers:** (a) 4 (b) 2 (c) 2 (d) 2 (e) $\frac{3}{4}$ (f) $\frac{2}{3}$ (g) 1 (h) 1 (i) NA