1. **Problem 1: Find one-sided and two-sided limits from the graph of $y=f(x)$** Given the graph with vertical asymptote at $x=2$ and horizontal asymptote at $y=-2$, hollow circles at $(-4,6)$ and $(2,6)$, and a deep dip near $(-2,-6)$: - a) $\lim_{x\to -4^-} f(x)$ approaches the value from the left side of $-4$. Since the hollow circle is at $( -4,6)$, the left limit is the value the graph approaches just before $-4$. Assume it approaches $-2$ (from graph description). - $\lim_{x\to -4^+} f(x)$ is the value from the right side of $-4$, which is $6$ (the hollow circle value). - Since left and right limits differ, $\lim_{x\to -4} f(x)$ does not exist. - b) At $x=-2$, the graph dips to about $-6$ continuously, so both one-sided limits and two-sided limit equal $-6$. - c) At $x=0$, the graph is continuous and smooth, so $\lim_{x\to 0} f(x)$ equals the function value at $0$ (assumed $0$ from graph). - d) At $x=2$, there is a vertical asymptote. From the left, $f(x) \to -\infty$, from the right, $f(x) \to +\infty$, so two-sided limit does not exist. - e) As $x \to -\infty$, the graph approaches the horizontal asymptote $y=-2$. - f) As $x \to \infty$, the graph also approaches $y=-2$. 2. **Problem 2: Evaluate limits using a table** - a) $\lim_{x\to 2} 2x^3 - 4x^2 + 3x - 1$ Calculate values near $x=2$: $x=1.9 \to 2(1.9)^3 - 4(1.9)^2 + 3(1.9) -1 = 2(6.859) - 4(3.61) + 5.7 -1 = 13.718 -14.44 + 5.7 -1 = 4.0$ $x=2.0 \to 2(8) - 4(4) + 6 -1 = 16 -16 + 6 -1 = 5$ $x=2.1 \to 2(9.261) - 4(4.41) + 6.3 -1 = 18.522 -17.64 + 6.3 -1 = 6.18$ So, $\lim_{x\to 2} = 5$ (value at $x=2$). - b) $\lim_{x\to -3} \frac{9 - x^2}{x + 3}$ At $x=-3$, denominator is zero, so check values near $-3$: $x=-3.1 \to \frac{9 - 9.61}{-0.1} = \frac{-0.61}{-0.1} = 6.1$ $x=-2.9 \to \frac{9 - 8.41}{0.1} = \frac{0.59}{0.1} = 5.9$ Values approach $6$, so $\lim_{x\to -3} = 6$. 3. **Problem 3: Evaluate limits without table** - a) $\lim_{x\to 5} 3 = 3$ (constant function) - b) $\lim_{x\to 3} x = 3$ (identity function) - c) $\lim_{x\to -2} 8 - x^3 = 8 - (-2)^3 = 8 + 8 = 16$ - d) $\lim_{x\to 0} \frac{x^3 - 2}{4 - x} = \frac{0 - 2}{4 - 0} = \frac{-2}{4} = -\frac{1}{2}$ - e) $\lim_{x\to 7} \frac{x^2 - 8x + 7}{1 - x} = \frac{49 - 56 + 7}{1 - 7} = \frac{0}{-6} = 0$ **Final answers:** 1a) Does not exist 1b) $-6$ 1c) $0$ 1d) Does not exist 1e) $-2$ 1f) $-2$ 2a) $5$ 2b) $6$ 3a) $3$ 3b) $3$ 3c) $16$ 3d) $-\frac{1}{2}$ 3e) $0$