1. **Problem Statement:** Estimate the limits using tables of values and evaluate given limits. 2. **Important formulas and rules:** - For limits involving logarithms and exponentials, use properties like $\lim_{x \to a} \frac{\ln x - \ln a}{x - a} = \frac{1}{a}$. - For trigonometric limits, use standard limits such as $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$. - For polynomial limits, direct substitution works unless it leads to indeterminate forms. - For limits of the form $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$. --- ### Question 2 (a) Estimate $\lim_{x \to 4} \frac{\ln x - \ln 4}{x - 4}$ - This is the definition of the derivative of $\ln x$ at $x=4$. - Formula: $\lim_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$. - Derivative: $\frac{d}{dx} \ln x = \frac{1}{x}$. - So, limit = $\frac{1}{4} = 0.25$. (b) Estimate $\lim_{p \to -1} \frac{1 + p^9}{1 + p^{15}}$ - Substitute $p = -1$: - Numerator: $1 + (-1)^9 = 1 - 1 = 0$ - Denominator: $1 + (-1)^{15} = 1 - 1 = 0$ - Indeterminate form $\frac{0}{0}$, use values close to $-1$: - For $p = -1.01$, numerator $\approx 1 + (-1.01)^9 \approx 1 - 1.093 = -0.093$ - Denominator $\approx 1 + (-1.01)^{15} \approx 1 - 1.160 = -0.160$ - Ratio $\approx \frac{-0.093}{-0.160} = 0.58$ - For $p = -0.99$, ratio $\approx 0.58$ - So limit $\approx 0.58$. (c) Estimate $\lim_{\theta \to 0} \frac{\sin 3\theta}{\tan 2\theta}$ - Use small angle approximations: - $\sin 3\theta \approx 3\theta$ - $\tan 2\theta \approx 2\theta$ - Limit $= \frac{3\theta}{2\theta} = \frac{3}{2} = 1.5$. (d) Estimate $\lim_{t \to 0} \frac{5^t - 1}{t}$ - Use formula $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$. - Here $a=5$, so limit $= \ln 5 \approx 1.609$. (e) Estimate $\lim_{x \to 0^+} x^x$ - Rewrite as $e^{x \ln x}$. - As $x \to 0^+$, $x \ln x \to 0$. - So limit $= e^0 = 1$. (f) Estimate $\lim_{x \to 0^+} x^2 \ln x$ - As $x \to 0^+$, $\ln x \to -\infty$ but $x^2 \to 0$. - Use substitution or L'Hopital's rule: - Limit is $0$ because $x^2$ goes to zero faster than $\ln x$ diverges. --- ### Question 3 (a) $\lim_{x \to 3} (5x^3 - 3x^2 + x - 6)$ - Direct substitution: - $5(3)^3 - 3(3)^2 + 3 - 6 = 5(27) - 3(9) + 3 - 6 = 135 - 27 + 3 - 6 = 105$. (b) $\lim_{u \to -2} \sqrt{u^4 + 3u + 6}$ - Substitute $u = -2$: - $(-2)^4 + 3(-2) + 6 = 16 - 6 + 6 = 16$. - Square root $= 4$. (c) $\lim_{x \to 5} \frac{x^2 - 6x + 5}{x - 5}$ - Factor numerator: - $x^2 - 6x + 5 = (x - 5)(x - 1)$. - Cancel $(x - 5)$: - Limit $= \lim_{x \to 5} (x - 1) = 4$. (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 form, factor denominator: - $x^2 - x - 12 = (x - 4)(x + 3)$ - Numerator: $x(x + 3)$ - Cancel $(x + 3)$: - Limit $= \lim_{x \to -3} \frac{x}{x - 4} = \frac{-3}{-7} = \frac{3}{7}$. (e) $\lim_{x \to 5} \frac{x^2 - 5x + 6}{x - 5}$ - Factor numerator: - $(x - 2)(x - 3)$ - 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 $= 16 - 4 - 12 = 0$ - Check factor denominator: - $(x - 4)(x + 3)$ - Cancel $(x - 4)$ if numerator divisible by $(x - 4)$? - Numerator $= x(x + 3)$, so no $(x - 4)$ factor. - Limit does not exist (infinite). (g) $\lim_{t \to -3} \frac{t^2 - 9}{2t^2 + 7t + 3}$ - Factor numerator: $(t - 3)(t + 3)$ - Factor denominator: $(2t + 1)(t + 3)$ - Cancel $(t + 3)$: - Limit $= \lim_{t \to -3} \frac{t - 3}{2t + 1} = \frac{-3 - 3}{2(-3) + 1} = \frac{-6}{-6 + 1} = \frac{-6}{-5} = \frac{6}{5}$. (h) $\lim_{x \to -1} \frac{2x^2 + 3x + 1}{x^2 - 2x - 3}$ - Factor numerator: $(2x + 1)(x + 1)$ - Factor denominator: $(x - 3)(x + 1)$ - Cancel $(x + 1)$: - Limit $= \lim_{x \to -1} \frac{2x + 1}{x - 3} = \frac{2(-1) + 1}{-1 - 3} = \frac{-2 + 1}{-4} = \frac{-1}{-4} = \frac{1}{4}$. (i) $\lim_{h \to 0} \frac{(-5 + h)^2 - 25}{h}$ - Expand numerator: - $(-5 + h)^2 = 25 - 10h + h^2$ - Numerator $= 25 - 10h + h^2 - 25 = -10h + h^2$ - Divide by $h$: $\frac{-10h + h^2}{h} = -10 + h$ - Limit $= -10$. (j) $\lim_{h \to 0} \frac{(2 + h)^3 - 8}{h}$ - Expand numerator: - $(2 + h)^3 = 8 + 12h + 6h^2 + h^3$ - Numerator $= 8 + 12h + 6h^2 + h^3 - 8 = 12h + 6h^2 + h^3$ - Divide by $h$: $12 + 6h + h^2$ - Limit $= 12$. --- **Final answers:** 2(a) $0.25$ 2(b) $0.58$ 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) $12$