1. **State the problem:** We need to find the limits: a. $\lim_{x \to 0} 2g(x+1)$ b. $\lim_{x \to 0} [f(x) - g(x)]$ c. $\lim_{x \to 2} f(g(x))$ 2. **Recall limit properties:** - The limit of a sum/difference is the sum/difference of the limits. - The limit of a constant times a function is the constant times the limit of the function. - The limit of a composition $f(g(x))$ as $x \to a$ is $f(\lim_{x \to a} g(x))$ if $f$ is continuous at that limit. 3. **Analyze $g(x)$ from the graph:** - $g(x)$ is a V-shaped piecewise linear function with vertex at $(0,4)$. - At $x=0$, $g(0) = 4$. - For $x$ near 1, $g(1)$ is on the line descending from $(0,4)$ to $(3,0)$. - The slope from $(0,4)$ to $(3,0)$ is $\frac{0-4}{3-0} = -\frac{4}{3}$. - So $g(1) = 4 + (-\frac{4}{3})(1-0) = 4 - \frac{4}{3} = \frac{8}{3} \approx 2.67$. 4. **Calculate each limit:** a. $\lim_{x \to 0} 2g(x+1) = 2 \lim_{x \to 0} g(x+1) = 2 g(1)$ (since $g$ is continuous at 1) $$= 2 \times \frac{8}{3} = \frac{16}{3} \approx 5.33$$ b. $\lim_{x \to 0} [f(x) - g(x)] = \lim_{x \to 0} f(x) - \lim_{x \to 0} g(x)$ From the graph: - $f(0) \approx 0$ - $g(0) = 4$ So, $$= 0 - 4 = -4$$ c. $\lim_{x \to 2} f(g(x)) = f\left(\lim_{x \to 2} g(x)\right)$ (assuming $f$ continuous at that point) From $g(x)$ graph: - For $x$ near 2, $g(2)$ lies on the line from $(0,4)$ to $(3,0)$. - Slope is $-\frac{4}{3}$, so $$g(2) = 4 + (-\frac{4}{3})(2-0) = 4 - \frac{8}{3} = \frac{4}{3} \approx 1.33$$ From $f(x)$ graph: - At $x \approx 1.33$, $f(x)$ is between the minimum at $-1$ and the peak at $1$. - The graph rises from $-4$ at $x=-1$ to $3$ at $x=1$, so at $x=1.33$ it is slightly less than 3, roughly about 2. So, $$\lim_{x \to 2} f(g(x)) \approx 2$$ **Final answers:** $$\lim_{x \to 0} 2g(x+1) = \frac{16}{3}$$ $$\lim_{x \to 0} [f(x) - g(x)] = -4$$ $$\lim_{x \to 2} f(g(x)) \approx 2$$