1. **Problem Statement:** Determine the correct answers for the limit problems given, including conditions for limit existence, evaluating limits from tables and graphs, and calculating limits from piecewise functions. 2. **Key Concept:** The limit of a function $\lim_{x \to a} f(x) = L$ exists if and only if the left-hand limit $\lim_{x \to a^-} f(x) = L$ and the right-hand limit $\lim_{x \to a^+} f(x) = L$ are equal. 3. **Question 1:** Under what condition does $\lim_{x \to a} f(x)$ exist and equal $L$? - Correct answer: C. The left-hand limit $\lim_{x \to a^-} f(x) = L$ and the right-hand limit $\lim_{x \to a^+} f(x) = L$. 4. **Question 2:** Function has a hole at $x=2$ with $f(2)=3$ but approaches 5 from both sides. - The limit depends on the values approaching $x=2$, not the function value at $x=2$. - Correct answer: C. $\lim_{x \to 2} f(x) = 5$. 5. **Question 3:** Table near $x=0$ shows $g(x)$ values approaching 4 from both sides. - Left-hand limit $\approx 4.0$, right-hand limit $\approx 3.95$ close to 4. - Correct answer: C. $\lim_{x \to 0} f(x) = 4$. 6. **Question 4:** Graph approaches infinity on both sides as $x \to -3$. - Limit is infinite. - Correct answer: C. $\lim_{x \to -3} f(x) = \infty$. 7. **Question B (First function):** $f(x) = \frac{x^2 - 2x - 35}{x + 5}$ near $x = -5$. - Factor numerator: $x^2 - 2x - 35 = (x - 7)(x + 5)$. - Simplify: $f(x) = \frac{(x - 7)(x + 5)}{x + 5} = x - 7$, for $x \neq -5$. - Calculate limits: - $\lim_{x \to -5^+} f(x) = (-5) - 7 = -12$ - $\lim_{x \to -5^-} f(x) = (-5) - 7 = -12$ - $\lim_{x \to -5} f(x) = -12$ - $f(-5)$ is undefined (division by zero). 8. **Question B (Second function):** Piecewise function $$f(x) = \begin{cases} \frac{x^2}{3x - 36}, & x > 6 \\ x^2 - 5x - 6, & x \leq 6 \end{cases}$$ - Calculate limits at $x=6$: - For $x > 6$, $f(x) = \frac{x^2}{3x - 36}$ - $\lim_{x \to 6^+} f(x) = \frac{6^2}{3(6) - 36} = \frac{36}{18 - 36} = \frac{36}{-18} = -2$ - For $x \leq 6$, $f(x) = x^2 - 5x - 6$ - $\lim_{x \to 6^-} f(x) = 6^2 - 5(6) - 6 = 36 - 30 - 6 = 0$ - Since left and right limits differ, $\lim_{x \to 6} f(x)$ does not exist. - $f(6) = 6^2 - 5(6) - 6 = 0$. 9. **Question C:** From the graph of $g(x)$ with vertical asymptotes at $x=0$ and $x=5$: - $\lim_{x \to 0^-} g(x) = -\infty$ (drops sharply) - $\lim_{x \to 0^+} g(x) = +\infty$ (rises sharply) - $\lim_{x \to 5^-} g(x) = +\infty$ (rises sharply) - $\lim_{x \to 5^+} g(x) = -\infty$ (drops sharply) - $\lim_{x \to 2^-} g(x) = g(2) = 1$ (solid dot at 2) - $\lim_{x \to 8^-} g(x) = 5$ (continuous increasing) - $g(2) = 1$, $g(8) = 5$ (from graph points) **Final answers:** 1. C 2. C 3. C 4. C $\lim_{x \to -5^+} f(x) = -12$, $\lim_{x \to -5^-} f(x) = -12$, $\lim_{x \to -5} f(x) = -12$, $f(-5)$ undefined $\lim_{x \to 6^+} f(x) = -2$, $\lim_{x \to 6^-} f(x) = 0$, $\lim_{x \to 6} f(x)$ does not exist, $f(6) = 0$ $\lim_{x \to 0^-} g(x) = -\infty$, $\lim_{x \to 0^+} g(x) = +\infty$, $\lim_{x \to 5^-} g(x) = +\infty$, $\lim_{x \to 5^+} g(x) = -\infty$, $\lim_{x \to 2^-} g(x) = 1$, $\lim_{x \to 8^-} g(x) = 5$, $g(2) = 1$, $g(8) = 5$