1. Stating the problem: We need to find various limits involving functions $f(x)$ and $g(x)$ using their graphs. 2. Analyze each limit step-by-step: --- **1.** $\lim_{x \to 1} [f(x) + g(x)]$ - From the graph of $f$, as $x \to 1$, $f(x) \to 1$ (peak point at $(1,1)$). - From the graph of $g$, as $x \to 1$, $g(x) \to 1$ (filled point at $(1,1)$, the value at the hollow point $(1,2)$ doesn’t affect limit). - Therefore, $\lim_{x \to 1} [f(x) + g(x)] = 1 + 1 = 2$. --- **2.** $\lim_{x \to 2} [f(x) + g(x)]$ - From $f$ graph near $x=2$, $f(x)$ is about $1.5$ (rising curve near 2). Estimate $f(2) \approx 1.5$. - From $g$ graph near $x=2$, $g(x)$ is approximately $3$ (graph rising upward). - So, limit is approximately $1.5 + 3 = 4.5$. --- **3.** $\lim_{x \to 0} f(x) g(x)$ - From $f$ graph near zero, there is a hole below zero but filled point at $f(0)=1$. - For limit, consider $f(x)$ approaching the hole value, around $0$ (since the curve dips below 0 at 0). - From $g$ graph at zero, $g(0) = 0$. - So, $\lim_{x \to 0} f(x) g(x) = 0 \times 0 = 0$. --- **4.** $\lim_{x \to 0} \frac{f(x)}{g(x)}$ - From above, near zero, $f(x)$ approaches $0$ (hole value) and $g(x)$ approaches $0$ (graph passes through origin). - Here, evaluate the ratio by left and right limits: - From $f$ graph approaching $0$ from left, $f(x) \to 0^-$ (slightly below zero), from right $f(x) \to 0^+$ (slightly above zero). - From $g(x)$ near zero is smooth crossing 0, so $g(x) \to 0$ linearly. - Approximating, the ratio approaches a finite number since they approach zero linearly and both are approx proportional. - Closer estimate suggests $\lim_{x \to 0} \frac{f(x)}{g(x)} \approx 1$ --- **5.** $\lim_{x \to 1} \sqrt{3 + f(x)}$ - Since $f(1) = 1$, inside the root is $3+1=4$. - Limits preserve continuity inside square root where defined. - So, $\lim_{x \to 1} \sqrt{3 + f(x)} = \sqrt{4} = 2$. --- **Final answers:** $$ \lim_{x \to 1} [f(x) + g(x)] = 2 $$ $$ \lim_{x \to 2} [f(x) + g(x)] = 4.5 $$ $$ \lim_{x \to 0} f(x) g(x) = 0 $$ $$ \lim_{x \to 0} \frac{f(x)}{g(x)} = 1 $$ $$ \lim_{x \to 1} \sqrt{3 + f(x)} = 2 $$