1. **Problem 1: One-sided and two-sided limits from the graph** Given the graph of $y=f(x)$ with a horizontal asymptote at $y=-2$ and a vertical asymptote at $x=2$, we find the limits at specified points. - **Important:** The two-sided limit $\lim_{x \to a} f(x)$ exists only if the left-hand limit $\lim_{x \to a^-} f(x)$ and right-hand limit $\lim_{x \to a^+} f(x)$ both exist and are equal. **a) $\lim_{x \to -4} f(x)$:** - From the graph, as $x$ approaches $-4$ from left and right, $f(x)$ approaches approximately $4$. - So, $\lim_{x \to -4^-} f(x) = 4$ and $\lim_{x \to -4^+} f(x) = 4$. - Therefore, $\lim_{x \to -4} f(x) = 4$. **b) $\lim_{x \to -2} f(x)$:** - From the graph, as $x$ approaches $-2$ from left and right, $f(x)$ approaches approximately $-8$. - So, $\lim_{x \to -2^-} f(x) = -8$ and $\lim_{x \to -2^+} f(x) = -8$. - Therefore, $\lim_{x \to -2} f(x) = -8$. **c) $\lim_{x \to 0} f(x)$:** - From the graph, as $x$ approaches $0$ from left and right, $f(x)$ approaches approximately $-7$. - So, $\lim_{x \to 0^-} f(x) = -7$ and $\lim_{x \to 0^+} f(x) = -7$. - Therefore, $\lim_{x \to 0} f(x) = -7$. **d) $\lim_{x \to 2} f(x)$:** - The graph has a vertical asymptote at $x=2$. - As $x \to 2^-$, $f(x) \to -\infty$ (steeply down). - As $x \to 2^+$, $f(x) \to \infty$ (rises sharply). - Since left and right limits are not equal, $\lim_{x \to 2} f(x)$ does not exist. **e) $\lim_{x \to -\infty} f(x)$:** - As $x \to -\infty$, from the graph, $f(x)$ approaches approximately $8$. **f) $\lim_{x \to \infty} f(x)$:** - As $x \to \infty$, $f(x)$ approaches the horizontal asymptote $y = -2$. 2. **Problem 2: Evaluate limits using a table** **a) $\lim_{x \to 2} 2x^3 - 4x^2 + 3x - 1$** - Substitute $x=2$ directly: $$2(2)^3 - 4(2)^2 + 3(2) - 1 = 2(8) - 4(4) + 6 - 1 = 16 - 16 + 6 - 1 = 5$$ **b) $\lim_{x \to -3} \frac{9 - x^2}{x + 3}$** - Direct substitution gives denominator $0$, so use factorization: $$9 - x^2 = (3 - x)(3 + x)$$ - So, $$\frac{9 - x^2}{x + 3} = \frac{(3 - x)(3 + x)}{x + 3}$$ - Cancel common factor $3 + x$ (which is $x + 3$): $$\frac{(3 - x)\cancel{(3 + x)}}{\cancel{x + 3}} = 3 - x$$ - Now substitute $x = -3$: $$3 - (-3) = 6$$ 3. **Problem 3: Evaluate limits without a table** **a) $\lim_{x \to 5} 3$** - Constant function, limit is $3$. **b) $\lim_{x \to 3} x$** - Direct substitution: $3$. **c) $\lim_{x \to 2} 8 - x^3$** - Substitute $x=2$: $$8 - 2^3 = 8 - 8 = 0$$ **d) $\lim_{x \to 0} \frac{x^3 - 2}{4 - x}$** - Substitute $x=0$: $$\frac{0 - 2}{4 - 0} = \frac{-2}{4} = -\frac{1}{2}$$ **e) $\lim_{x \to 7} \frac{x^2 - 8x + 7}{1 - x}$** - Factor numerator: $$x^2 - 8x + 7 = (x - 7)(x - 1)$$ - So, $$\frac{(x - 7)(x - 1)}{1 - x}$$ - Note $1 - x = -(x - 1)$, so: $$\frac{(x - 7)(x - 1)}{-(x - 1)} = -(x - 7)$$ - Cancel $x - 1$: $$\frac{(x - 7)\cancel{(x - 1)}}{-\cancel{(x - 1)}} = -(x - 7)$$ - Substitute $x=7$: $$-(7 - 7) = 0$$ 4. **Problem 4: Values of $c$ where limits do not exist** **a) $\lim_{x \to c} \sqrt{5 - x}$** - The expression under the square root must be non-negative: $$5 - c \geq 0 \implies c \leq 5$$ - For $c > 5$, the function is not defined, so limit does not exist. **b) $\lim_{x \to c} \frac{3}{x + 1}$** - Denominator cannot be zero: $$c + 1 \neq 0 \implies c \neq -1$$ - At $c = -1$, limit does not exist (vertical asymptote). **c) $\lim_{x \to c} \frac{4}{(x - 3)^2}$** - Denominator zero at $x=3$: $$c \neq 3$$ - At $c=3$, limit does not exist (vertical asymptote). **Final answers:** - 1a) 4 - 1b) -8 - 1c) -7 - 1d) Does not exist - 1e) 8 - 1f) -2 - 2a) 5 - 2b) 6 - 3a) 3 - 3b) 3 - 3c) 0 - 3d) -\frac{1}{2} - 3e) 0 - 4a) Limit does not exist for $c > 5$ - 4b) Limit does not exist for $c = -1$ - 4c) Limit does not exist for $c = 3$