1. **Problem Statement:** Estimate or evaluate the given limits using tables of values or algebraic simplification. --- ### Question 2: Use a table of values to estimate the limits. (a) $$\lim_{x \to 4} \frac{\ln x - \ln 4}{x - 4}$$ - This is the definition of the derivative of $\ln x$ at $x=4$. - The derivative formula: $$\frac{d}{dx} \ln x = \frac{1}{x}$$ - So, the limit equals $$\frac{1}{4} = 0.25$$. (b) $$\lim_{p \to -1} \frac{1 + p^9}{1 + p^{15}}$$ - Substitute $p = -1$ directly: - Numerator: $1 + (-1)^9 = 1 - 1 = 0$ - Denominator: $1 + (-1)^{15} = 1 - 1 = 0$ - Indeterminate form $\frac{0}{0}$, use values close to $-1$: - For $p = -1.01$, numerator $\approx 1 + (-1.01)^9$, denominator $\approx 1 + (-1.01)^{15}$. - Approximating shows limit approaches 1. (c) $$\lim_{\theta \to 0} \frac{\sin 3\theta}{\tan 2\theta}$$ - Use small angle approximations: $\sin x \approx x$, $\tan x \approx x$ as $x \to 0$. - So limit $\approx \frac{3\theta}{2\theta} = \frac{3}{2} = 1.5$. (d) $$\lim_{t \to 0} \frac{5^t - 1}{t}$$ - This is the definition of the derivative of $5^t$ at $t=0$. - Derivative: $$\frac{d}{dt} 5^t = 5^t \ln 5$$ - At $t=0$, value is $5^0 \ln 5 = 1 \times \ln 5 = \ln 5 \approx 1.609$. (e) $$\lim_{x \to 0^+} x^x$$ - Rewrite as $e^{x \ln x}$. - As $x \to 0^+$, $x \ln x \to 0$. - So limit is $e^0 = 1$. (f) $$\lim_{x \to 0^+} x^2 \ln x$$ - As $x \to 0^+$, $\ln x \to -\infty$, but $x^2 \to 0$. - Use substitution or L'Hôpital's rule to find limit is 0. --- ### Question 3: Evaluate the limits. (a) $$\lim_{x \to 3} (5x^3 - 3x^2 + x - 6)$$ - Direct substitution: $5(27) - 3(9) + 3 - 6 = 135 - 27 + 3 - 6 = 105$. (b) $$\lim_{u \to -2} \sqrt{u^4 + 3u + 6}$$ - Substitute $u = -2$: $\sqrt{16 - 6 + 6} = \sqrt{16} = 4$. (c) $$\lim_{x \to 5} \frac{x^2 - 6x + 5}{x - 5}$$ - Factor numerator: $(x-5)(x-1)$. - Cancel $(x-5)$, limit is $x-1$ at $x=5$, so $4$. (d) $$\lim_{x \to -3} \frac{x^2 + 3x}{x^2 - x - 12}$$ - Numerator: $(-3)^2 + 3(-3) = 9 - 9 = 0$. - Denominator: $9 + 3 - 12 = 0$. - Indeterminate form, factor: - Numerator: $x(x+3)$ - Denominator: $(x-4)(x+3)$ - Cancel $(x+3)$, limit is $\frac{x}{x-4}$ at $x=-3$, so $\frac{-3}{-7} = \frac{3}{7}$. (e) $$\lim_{x \to 5} \frac{x^2 - 5x + 6}{x - 5}$$ - Numerator factors: $(x-2)(x-3)$. - Substitute $x=5$: numerator $= (5-2)(5-3) = 3 \times 2 = 6$, denominator $= 0$. - Limit does not exist (division by zero). (f) $$\lim_{x \to 4} \frac{x^2 + 3x}{x^2 - x - 12}$$ - Substitute $x=4$: - Numerator: $16 + 12 = 28$ - Denominator: $16 - 4 - 12 = 0$ - Limit does not exist (division by zero). (g) $$\lim_{t \to -3} \frac{t^2 - 9}{2t^2 + 7t + 3}$$ - Numerator: $(t-3)(t+3)$ - Denominator: factor $2t^2 + 7t + 3 = (2t+3)(t+1)$ - Substitute $t=-3$: numerator $= 0$, denominator $= 2(9) - 21 + 3 = 18 - 21 + 3 = 0$ - Indeterminate form, cancel $(t+3)$: - Limit is $\lim_{t \to -3} \frac{t-3}{2t+3}$ - Substitute $t=-3$: $\frac{-6}{-3} = 2$. (h) $$\lim_{x \to -1} \frac{2x^2 + 3x + 1}{x^2 - 2x - 3}$$ - Numerator: $2(-1)^2 + 3(-1) + 1 = 2 - 3 + 1 = 0$ - Denominator: $1 + 2 - 3 = 0$ - Factor numerator: $(2x+1)(x+1)$ - Factor denominator: $(x-3)(x+1)$ - Cancel $(x+1)$, limit is $\lim_{x \to -1} \frac{2x+1}{x-3} = \frac{-2+1}{-1-3} = \frac{-1}{-4} = \frac{1}{4}$. (i) $$\lim_{h \to 0} \frac{(-5 + h)^2 - 25}{h}$$ - Expand numerator: $25 - 10h + h^2 - 25 = -10h + h^2$ - Expression: $\frac{-10h + h^2}{h} = -10 + h$ - Limit as $h \to 0$ is $-10$. (j) $$\lim_{h \to 0} \frac{(2 + h)^3 - 8}{h}$$ - Expand numerator: $(8 + 12h + 6h^2 + h^3) - 8 = 12h + 6h^2 + h^3$ - Expression: $\frac{12h + 6h^2 + h^3}{h} = 12 + 6h + h^2$ - Limit as $h \to 0$ is $12$. --- **Final answers:** 2(a) 0.25 2(b) 1 2(c) 1.5 2(d) $\ln 5 \approx 1.609$ 2(e) 1 2(f) 0 3(a) 105 3(b) 4 3(c) 4 3(d) $\frac{3}{7}$ 3(e) Does not exist 3(f) Does not exist 3(g) 2 3(h) $\frac{1}{4}$ 3(i) -10 3(j) 12