1. **Problem Statement:** Estimate the following limits using a table of values or direct substitution where possible. 2. **Recall:** The limit of a function $f(x)$ as $x$ approaches $a$ is the value $f(x)$ approaches as $x$ gets closer to $a$. --- ### Question 2 (a) $$\lim_{x \to 4} \frac{\ln x - \ln 4}{x - 4}$$ - This is the definition of the derivative of $\ln x$ at $x=4$. - Derivative of $\ln x$ is $\frac{1}{x}$. - So, limit = $\frac{1}{4} = 0.25$. (b) $$\lim_{p \to -1} \frac{1 + p^9}{1 + p^{15}}$$ - Substitute $p = -1$: - Numerator: $1 + (-1)^9 = 1 - 1 = 0$ - Denominator: $1 + (-1)^{15} = 1 - 1 = 0$ - Indeterminate form $\frac{0}{0}$, use algebra or L'Hôpital's rule. - Factor numerator and denominator: - Numerator: $1 + p^9 = (1+p)(1 - p + p^2 - p^3 + ... + p^8)$ - Denominator: $1 + p^{15} = (1+p)(1 - p + p^2 - ... + p^{14})$ - Cancel $(1+p)$: $$\lim_{p \to -1} \frac{1 - p + p^2 - p^3 + ... + p^8}{1 - p + p^2 - ... + p^{14}}$$ - Substitute $p = -1$: - Numerator sum: $1 - (-1) + (-1)^2 - ... + (-1)^8 = 1 + 1 + 1 + ...$ (9 terms alternating signs) - Calculate carefully: Terms: $1, +1, +1, +1, +1, +1, +1, +1, +1$ (actually alternating signs, so sum is 5) - Denominator sum similarly is 8. - So limit = $\frac{5}{8} = 0.625$. (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 of $a^t$ is $a^t \ln a$. - At $t=0$, derivative = $5^0 \ln 5 = 1 \times \ln 5 = \ln 5 \approx 1.609$. (e) $$\lim_{x \to 0^+} x^x$$ - Rewrite as $e^{x \ln x}$. - As $x \to 0^+$, $x \ln x \to 0$. - So limit = $e^0 = 1$. (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: - Let $y = x^2 \ln x = \frac{\ln x}{1/x^2}$. - Both numerator and denominator tend to infinity, apply L'Hôpital's: - Derivative numerator: $\frac{1}{x}$ - Derivative denominator: $-2/x^3$ - So limit becomes $\lim_{x \to 0^+} \frac{1/x}{-2/x^3} = \lim_{x \to 0^+} \frac{x^3}{-2x} = \lim_{x \to 0^+} \frac{x^2}{-2} = 0$. --- ### Question 3 (a) $$\lim_{x \to 3} (5x^3 - 3x^2 + x - 6)$$ - Polynomial, continuous everywhere. - Substitute $x=3$: - $5(27) - 3(9) + 3 - 6 = 135 - 27 + 3 - 6 = 105$. (b) $$\lim_{u \to -2} \sqrt{u^4 + 3u + 6}$$ - Substitute $u=-2$: - $(-2)^4 + 3(-2) + 6 = 16 - 6 + 6 = 16$. - Square root: $\sqrt{16} = 4$. (c) $$\lim_{x \to 5} \frac{x^2 - 6x + 5}{x - 5}$$ - Substitute $x=5$ numerator: $25 - 30 + 5 = 0$, denominator $0$. - Indeterminate form $0/0$. - Factor numerator: $(x-5)(x-1)$. - Cancel $(x-5)$: - Limit becomes $\lim_{x \to 5} (x-1) = 4$. (d) $$\lim_{x \to -3} \frac{x^2 + 3x}{x^2 - x - 12}$$ - Substitute $x=-3$ numerator: $9 - 9 = 0$, denominator: $9 + 3 - 12 = 0$. - Indeterminate $0/0$. - Factor numerator: $x(x+3)$. - Factor denominator: $(x-4)(x+3)$. - Cancel $(x+3)$: - Limit becomes $\lim_{x \to -3} \frac{x}{x-4} = \frac{-3}{-3-4} = \frac{-3}{-7} = \frac{3}{7}$. (e) $$\lim_{x \to 5} \frac{x^2 - 5x + 6}{x - 5}$$ - Substitute $x=5$ numerator: $25 - 25 + 6 = 6$, denominator $0$. - Limit does not exist (division by zero, numerator nonzero). (f) $$\lim_{x \to 4} \frac{x^2 + 3x}{x^2 - x - 12}$$ - Substitute $x=4$ numerator: $16 + 12 = 28$, denominator: $16 - 4 - 12 = 0$. - Denominator zero, numerator nonzero, limit does not exist. (g) $$\lim_{t \to -3} \frac{t^2 - 9}{2t^2 + 7t + 3}$$ - Substitute $t=-3$ numerator: $9 - 9 = 0$, denominator: $18 - 21 + 3 = 0$. - Indeterminate $0/0$. - Factor numerator: $(t-3)(t+3)$. - Factor denominator: $(2t+1)(t+3)$. - Cancel $(t+3)$: - Limit becomes $\lim_{t \to -3} \frac{t-3}{2t+1} = \frac{-3-3}{2(-3)+1} = \frac{-6}{-6+1} = \frac{-6}{-5} = \frac{6}{5}$. (h) $$\lim_{x \to -1} \frac{2x^2 + 3x + 1}{x^2 - 2x - 3}$$ - Substitute $x=-1$ numerator: $2 + (-3) + 1 = 0$, denominator: $1 + 2 - 3 = 0$. - Indeterminate $0/0$. - Factor numerator: $(2x+1)(x+1)$. - Factor denominator: $(x-3)(x+1)$. - Cancel $(x+1)$: - Limit becomes $\lim_{x \to -1} \frac{2x+1}{x-3} = \frac{2(-1)+1}{-1-3} = \frac{-2+1}{-4} = \frac{-1}{-4} = \frac{1}{4}$. (i) $$\lim_{h \to 0} \frac{(-5 + h)^2 - 25}{h}$$ - Expand numerator: $(-5 + h)^2 = 25 - 10h + h^2$. - Numerator: $25 - 10h + h^2 - 25 = -10h + h^2$. - Limit becomes $\lim_{h \to 0} \frac{-10h + h^2}{h} = \lim_{h \to 0} (-10 + h) = -10$. (j) $$\lim_{h \to 0} \frac{(2 + h)^3 - 8}{h}$$ - Expand numerator: $(2 + h)^3 = 8 + 12h + 6h^2 + h^3$. - Numerator: $8 + 12h + 6h^2 + h^3 - 8 = 12h + 6h^2 + h^3$. - Limit becomes $\lim_{h \to 0} \frac{12h + 6h^2 + h^3}{h} = \lim_{h \to 0} (12 + 6h + h^2) = 12$. --- **Final answers:** 2(a) 0.25 2(b) 0.625 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) $\frac{6}{5}$ 3(h) $\frac{1}{4}$ 3(i) -10 3(j) 12