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Section 8.7 #12: Evaluate the limit, using L'Hopital's rule if necessary: $$\displaystyle\lim_{x \rightarrow -2} \dfrac{x^2-3x-10}{x+2}.$$ Solution: Plugging in $x=-2$ yields indeterminate form $\dfrac{0}{0}$, so we may apply L'Hopital's rule: compute $$\displaystyle\lim_{x \rightarrow -2} \dfrac{x^2-3x-10}{x+2} \stackrel{L.H.}{=} \displaystyle\lim_{x \rightarrow -2} \dfrac{2x-3}{1} = -7.$$

Section 8.7 #20: Evaluate the limit, using L'Hopital's rule if necessary: $$\displaystyle\lim_{x \rightarrow 0} \dfrac{\sin(ax)}{\sin(bx)}, \quad a\neq 0, b \neq 0.$$ Solution: Plugging in $x=0$ yields indeterminate form $\dfrac{0}{0}$, so we may apply L'Hopital's rule: compute $$\displaystyle\lim_{x \rightarrow 0} \dfrac{\sin(ax)}{\sin(bx)} = \displaystyle\lim_{x \rightarrow 0} \dfrac{a\cos(ax)}{b\cos(bx)} = \dfrac{a\cos(0)}{b\cos(0)} = \dfrac{a}{b}.$$

Section 8.7 #23: Evaluate the limit, using L'Hopital's rule if necessary: $$\displaystyle\lim_{x \rightarrow \infty} \dfrac{5x^2+4x+7}{4x^2+5}.$$ Solution: "Plugging in" $x=\infty$ yields indeterminate form $\dfrac{\infty}{\infty}$, so we may apply L'Hopital's rule: compute $$\displaystyle\lim_{x \rightarrow \infty} \dfrac{5x^2+4x+7}{4x^2+5} \stackrel{L.H.}{=} \displaystyle\lim_{x \rightarrow \infty} \dfrac{10x+4}{8x} \stackrel{L.H.}{=} \displaystyle\lim_{x \rightarrow \infty} \dfrac{10}{8} = \dfrac{10}{8}.$$ Section 8.7 #28: Evaluate the limit, using L'Hopital's rule if necessary: $$\displaystyle\lim_{x \rightarrow \infty} \dfrac{x^3}{e^{x^2}}.$$ Solution: "Plugging in" $x=\infty$ yields indeterminate form $\dfrac{\infty}{\infty}$ and so we may apply L'Hopital's rule (note that we will apply it successively for the same reason): compute $$\begin{array}{ll} \displaystyle\lim_{x \rightarrow \infty} \dfrac{x^3}{e^{x^2}} &\stackrel{L.H.}{=} \displaystyle\lim_{x \rightarrow \infty} \dfrac{3x^2}{2xe^{x^2}} \\ &\stackrel{L.H.}{=} \displaystyle\lim_{x \rightarrow \infty} \dfrac{6x}{2e^{x^2} + 4x^2e^{x^2}} \\ &\stackrel{L.H.}{=} \displaystyle\lim_{x \rightarrow \infty} \dfrac{2}{4xe^{x^2}+8xe^{x^2}+8x^3e^{x^2}} \\ &= 0. \end{array}$$

Section 8.8 #17: Determine if the improper integral converges or diverges. Evaluate it if it converges: $$\displaystyle\int_1^{\infty} \dfrac{1}{x^3} \mathrm{d}x.$$ Solution: Compute $$\begin{array}{ll} \displaystyle\int_1^{\infty} \dfrac{1}{x^3} \mathrm{d}x &= \displaystyle\lim_{b \rightarrow \infty}\int_1^b x^{-3} \mathrm{d}x \\ &=\displaystyle\lim_{b \rightarrow \infty} -\dfrac{1}{2} x^{-2} \Bigg|_1^b \\ &= -\dfrac{1}{2} \displaystyle\lim_{b \rightarrow \infty} \left( \dfrac{1}{b^2} - 1 \right) \\ &= \left( -\dfrac{1}{2} \right) (-1) \\ &= \dfrac{1}{2}. \end{array}$$ Therefore we have shown that the improper integral converges.

Section 8.8 #20: Determine if the improper integral converges or diverges. Evaluate it if it converges: $$\displaystyle\int_1^{\infty} \dfrac{4}{\sqrt[4]{x}} \mathrm{d}x.$$ Solution: Compute $$\begin{array}{ll} \displaystyle\int_1^{\infty} \dfrac{4}{\sqrt[4]{x}} \mathrm{d}x &= 4 \displaystyle\lim_{b \rightarrow \infty} \displaystyle\int_1^b x^{-\frac{1}{4}} \mathrm{d}x \\ &= 4 \displaystyle\lim_{b \rightarrow \infty} \dfrac{x^{\frac{3}{4}}}{\frac{3}{4}} \Bigg|_1^b \\ &=\dfrac{16}{3} \displaystyle\lim_{b \rightarrow \infty} \left( b^{\frac{3}{4}} - 1 \right) \\ &= \infty, \end{array}$$ since $b^{\frac{3}{4}} \rightarrow \infty$ as $b \rightarrow \infty$. Therefore we have shown that the integral diverges.

Section 8.8 #38: Check whether the improper integral converges or diverges. Evaluate it if it converges: $$\displaystyle\int_0^e \ln(x^2) \mathrm{d}x.$$ Solution: This is an improper integral because the lower limit of integration, zero, is an asymptote of $\ln(x^2)$. Also recall that $\ln(x^n)n\ln(x)$ and $\displaystyle\int \ln(x) \mathrm{d}x = x\ln(x)-x+C$. Therefore compute $$\begin{array}{ll} \displaystyle\int_0^e \ln(x^2) \mathrm{d}x &= 2 \displaystyle\lim_{a \rightarrow 0^+} \int_a^e \ln(x) \mathrm{d}x \\ &= 2 \displaystyle\lim_{a \rightarrow 0^+} x \ln(x) - x \Bigg|_a^e \\ &= 2 \displaystyle\lim_{a \rightarrow 0^+} ( (e\ln(e)-e) - (a \ln(a)-a) ). \end{array}$$ Since plugging in $a=0$ into $a\ln(a)-a \rightarrow 0 \cdot (-\infty)$ is an indeterminate form, we may apply L'Hopital's rule to determine the limit (we will write the expression $a\ln(a)$ as $\frac{\ln(a)}{\frac{1}{a}}$): $$\begin{array}{ll} \displaystyle\lim_{a \rightarrow 0^+} a \ln(a) &\stackrel{L.H.}{=} \displaystyle\lim_{a \rightarrow 0^+} \dfrac{\frac{\mathrm{d}}{\mathrm{d}a} \ln(a)}{\frac{\mathrm{d}}{\mathrm{d}a} \frac{1}{a} } \\ &= \displaystyle\lim_{a \rightarrow 0^+} \dfrac{\frac{1}{a}}{-\frac{1}{a^2}} \\ &= \displaystyle\lim_{a \rightarrow 0^+} \dfrac{-a^2}{a} \\ &= \displaystyle\lim_{a \rightarrow 0^+} -a \\ &= 0. \end{array}$$ Therefore we have shown the the integral converges and $$\displaystyle\int_0^e \ln(x^2) \mathrm{d}x = 2 (e \ln(e)-e) = 2 (e-e) = 0.$$

Section 9.1 #18: Simplify the ratio of factorials:
$$\dfrac{n!}{(n+2)!}.$$ Solution: Recall that $m!=m(m-1)(m-2) \ldots 3 \cdot 2 \cdot 1$. Therefore, compute $$\dfrac{n!}{(n+2)!} = \dfrac{n!}{(n+2)(n+1)n!} = \dfrac{1}{(n+2)(n+1)}.$$