### Problem 31 Find (a) $\lim_{x \to 3^-} f(x)$ (b) $\lim_{x \to 3^+} f(x)$ (c) $\lim_{x \to 3} f(x)$ Given $$ f(x) = \begin{cases} x - 1, & x \leq 3 \\ 3x - 7, & x > 3 \end{cases} $$ 1. To find $\lim_{x \to 3^-} f(x)$, use the expression for $x \leq 3$: $$ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x - 1) = 3 - 1 = 2 $$ 2. For $\lim_{x \to 3^+} f(x)$, use the expression for $x > 3$: $$ \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (3x - 7) = 3 \times 3 - 7 = 9 - 7 = 2 $$ 3. Since the left and right limits at $x = 3$ are equal, the two-sided limit exists and equals 2: $$ \lim_{x \to 3} f(x) = 2 $$ ### Problem 8 Find $\lim_{x \to -2} \frac{x^3 + 8}{x^2 + 2}$ 1. Direct substitution leads to denominator zero: $$ (-2)^2 + 2 = 4 + 2 = 6 \neq 0 $$ Actually denominator at $x=-2$ is 6, no zero issue. 2. Calculate numerator: $$ (-2)^3 + 8 = -8 + 8 = 0 $$ 3. The limit is numerator over denominator: $$ \lim_{x \to -2} \frac{x^3 + 8}{x^2 + 2} = \frac{0}{6} = 0 $$ ### Problem 11 Find $\lim_{x \to -1^-} \frac{2x^2 + x - 1}{x + 1}$ 1. Substitute $x = -1$ into denominator: $$ -1 + 1 = 0 $$ Denominator zero, check numerator: $$ 2(-1)^2 + (-1) - 1 = 2 - 1 - 1 = 0 $$ We have an indeterminate form $0/0$, so factor: 2. Factor numerator: $$ 2x^2 + x - 1 = (2x - 1)(x + 1) $$ 3. Simplify limit expression: $$ \frac{2x^2 + x - 1}{x + 1} = \frac{(2x - 1)(x + 1)}{x + 1} = 2x - 1, \quad x \neq -1 $$ 4. Now evaluate limit: $$ \lim_{x \to -1^-} 2x - 1 = 2(-1) - 1 = -2 -1 = -3 $$ ### Problem 28 Find $\lim_{x \to 3^-} \frac{1}{|x - 3|}$ 1. For $x \to 3^-$, $x - 3$ is negative, so: $$ |x - 3| = -(x - 3) = 3 - x $$ 2. The limit becomes: $$ \lim_{x \to 3^-} \frac{1}{3 - x} $$ 3. As $x \to 3^-$, $3 - x \to 0^+$, so reciprocal tends to $+\infty$. ### Problem 40 Given $$ f(x) = \begin{cases} \frac{x^2 - 9}{x + 3}, & x \neq -3 \\ k, & x = -3 \end{cases} $$ (a) Find $k$ so $f$ is continuous at $x = -3$. 1. Factor numerator: $$ x^2 - 9 = (x - 3)(x + 3) $$ 2. For $x \neq -3$: $$ f(x) = \frac{(x - 3)(x + 3)}{x + 3} = x - 3 $$ 3. Find $\lim_{x \to -3} f(x) = \lim_{x \to -3} (x - 3) = -3 - 3 = -6$ 4. For continuity: $$ k = f(-3) = -6 $$ (b) Express $f(x)$ as a polynomial with $k = -6$: Since $f(x) = x - 3$ for $x \neq -3$ and $f(-3) = -6$, define $$ F(x) = x - 3 $$ which is a polynomial continuous everywhere including $x = -3$. --- Final answers: (a) $\lim_{x \to 3^-} f(x) = 2$ (b) $\lim_{x \to 3^+} f(x) = 2$ (c) $\lim_{x \to 3} f(x) = 2$ Problem 8 limit = 0 Problem 11 limit = -3 Problem 28 limit = +\infty Problem 40a: $k = -6$ Problem 40b: $f(x) = x - 3$