1. Find the following limits: **a)** $\lim_{x \to -2} \sqrt{4x^2 - 3}$ Step 1: Substitute $x = -2$ directly: $$\sqrt{4(-2)^2 - 3} = \sqrt{4 \times 4 - 3} = \sqrt{16 - 3} = \sqrt{13}$$ Answer: $\sqrt{13}$ **b)** $\lim_{x \to -5} \frac{x^2 + 3x - 10}{x + 5}$ Step 1: Factor numerator: $$x^2 + 3x - 10 = (x + 5)(x - 2)$$ Step 2: Simplify expression: $$\frac{(x + 5)(x - 2)}{x + 5} = x - 2, \quad x \neq -5$$ Step 3: Substitute $x = -5$: $$-5 - 2 = -7$$ Answer: $-7$ **c)** $\lim_{x \to -1} \frac{\sqrt{x^2 + 8} - 3}{x + 1}$ Step 1: Direct substitution gives $0/0$, so rationalize numerator: $$\frac{\sqrt{x^2 + 8} - 3}{x + 1} \times \frac{\sqrt{x^2 + 8} + 3}{\sqrt{x^2 + 8} + 3} = \frac{x^2 + 8 - 9}{(x + 1)(\sqrt{x^2 + 8} + 3)} = \frac{x^2 - 1}{(x + 1)(\sqrt{x^2 + 8} + 3)}$$ Step 2: Factor numerator: $$x^2 - 1 = (x + 1)(x - 1)$$ Step 3: Cancel $(x + 1)$: $$\frac{(x + 1)(x - 1)}{(x + 1)(\sqrt{x^2 + 8} + 3)} = \frac{x - 1}{\sqrt{x^2 + 8} + 3}$$ Step 4: Substitute $x = -1$: $$\frac{-1 - 1}{\sqrt{(-1)^2 + 8} + 3} = \frac{-2}{\sqrt{1 + 8} + 3} = \frac{-2}{3 + 3} = \frac{-2}{6} = -\frac{1}{3}$$ Answer: $-\frac{1}{3}$ **d)** $\lim_{x \to -1} \frac{x^2 - 1}{|x| - 1}$ Step 1: Factor numerator: $$x^2 - 1 = (x - 1)(x + 1)$$ Step 2: Evaluate denominator at $x \to -1$: $$|x| - 1 = |-1| - 1 = 1 - 1 = 0$$ Step 3: Substitute $x = -1$ in numerator: $$( -1 - 1)(-1 + 1) = (-2)(0) = 0$$ Step 4: Use left and right limits: - For $x \to -1^-$, $|x| = -x$, so denominator $= -x - 1$. - For $x \to -1^+$, $|x| = x$, so denominator $= x - 1$. Step 5: Calculate left limit: $$\lim_{x \to -1^-} \frac{(x - 1)(x + 1)}{-x - 1}$$ At $x = -1$, numerator $= 0$, denominator $= -(-1) - 1 = 1 - 1 = 0$, indeterminate. Try values near $-1$ from left: Numerator near $-1$ is negative times zero, denominator near zero positive. Step 6: Calculate right limit: $$\lim_{x \to -1^+} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to -1^+} (x + 1) = 0$$ Step 7: Since left and right limits differ, limit does not exist. Answer: Limit does not exist. **e)** $\lim_{u \to 0} \frac{\tan 2u}{u}$ Step 1: Use limit property $\lim_{x \to 0} \frac{\tan x}{x} = 1$ Step 2: Rewrite: $$\lim_{u \to 0} \frac{\tan 2u}{u} = \lim_{u \to 0} 2 \cdot \frac{\tan 2u}{2u} = 2 \times 1 = 2$$ Answer: $2$ **f)** $\lim_{x \to \infty} \frac{2 + \sqrt{x}}{2 - \sqrt{x}}$ Step 1: Divide numerator and denominator by $\sqrt{x}$: $$\frac{\frac{2}{\sqrt{x}} + 1}{\frac{2}{\sqrt{x}} - 1}$$ Step 2: As $x \to \infty$, $\frac{2}{\sqrt{x}} \to 0$, so limit becomes: $$\frac{0 + 1}{0 - 1} = \frac{1}{-1} = -1$$ Answer: $-1$ **g)** $\lim_{\theta \to -\infty} \frac{\cos \theta}{3\theta}$ Step 1: $\cos \theta$ oscillates between $-1$ and $1$. Step 2: Denominator $3\theta \to -\infty$. Step 3: Since numerator is bounded and denominator grows without bound, limit is 0. Answer: $0$ 2. Find the asymptotes and graph the functions: **a)** $y = \frac{1}{2x + 4}$ Step 1: Vertical asymptote where denominator is zero: $$2x + 4 = 0 \Rightarrow x = -2$$ Step 2: Horizontal asymptote as $x \to \pm \infty$: $$y \to 0$$ Answer: Vertical asymptote $x = -2$, horizontal asymptote $y = 0$ **b)** $y = \frac{x + 3}{x + 2}$ Step 1: Vertical asymptote where denominator zero: $$x + 2 = 0 \Rightarrow x = -2$$ Step 2: Horizontal asymptote by dividing numerator and denominator by $x$: $$y = \frac{1 + \frac{3}{x}}{1 + \frac{2}{x}} \to \frac{1 + 0}{1 + 0} = 1$$ Answer: Vertical asymptote $x = -2$, horizontal asymptote $y = 1$ **c)** $f(x) = \frac{x^2 - 3}{2x - 4}$ Step 1: Vertical asymptote where denominator zero: $$2x - 4 = 0 \Rightarrow x = 2$$ Step 2: Find oblique (slant) asymptote by polynomial division: Divide $x^2 - 3$ by $2x - 4$: - Leading term: $x^2 / 2x = \frac{x}{2}$ - Multiply back: $\frac{x}{2} \times (2x - 4) = x^2 - 2x$ - Subtract: $(x^2 - 3) - (x^2 - 2x) = 2x - 3$ - Next term: $2x / 2x = 1$ - Multiply back: $1 \times (2x - 4) = 2x - 4$ - Subtract: $(2x - 3) - (2x - 4) = 1$ So quotient is $\frac{x}{2} + 1$ with remainder $1$. Step 3: Oblique asymptote: $$y = \frac{x}{2} + 1$$ Answer: Vertical asymptote $x = 2$, oblique asymptote $y = \frac{x}{2} + 1$