112年政大風險管理精算組碩士班-微積分詳解

 國立政治大學112學年度碩士班暨碩士在職專班

考試科目:微積分
系所別:風險管理與保險系精算科學組

解答: $$\textbf{(a)}\;  \int{xdx\over \sqrt{8-2x^2-x^4}} = \int{xdx\over \sqrt{9-(x^2+1)^2}} = \int{xdx\over 3\sqrt{1-({x^2+1\over 3})^2}} \\\qquad \text{Let }u ={x^2+1\over 3} \Rightarrow du={2\over 3}xdx \Rightarrow xdx ={3\over 2}du \Rightarrow \int{xdx\over 3\sqrt{1-({x^2+1\over 3})^2}}= \int {{3\over 2}du \over 3\sqrt{1-u^2}} \\\qquad ={1\over 2}\int {1\over \sqrt{1-u^2}}du ={1\over 2}\sin^{-1} u+c =\bbox[red, 2pt]{{1\over 2}\sin^{-1}{x^2+1\over 3}+c} \\\textbf{(b)}\;  \int_1^2  {x^2+4\over x^4+3x^3+2x^2}\,dx =\int_1^2  {-3\over x }+{5\over x+1} +{-2\over x+2}+{2\over x^2}\,dx\\\qquad =\left. \left[ -3\ln x+5\ln(x+1)-2\ln(x+2)-{2\over x} \right] \right|_1^2= \bbox[red, 2pt]{7\ln 3-12\ln 2+1}\\ \textbf{(c)}\;\cases{u=x^2\\ dv=e^{-x}dx} \Rightarrow \cases{du =2xdx\\ v=-e^{-x}} \Rightarrow \int  x^2e^{-x}\,dx =-x^2e^{-x}+\int 2xe^{-x}\,dx \\\qquad =-x^2e^{-x}-2xe^{-x}-2e^{-x}+c \Rightarrow \int_0^\infty  x^2e^{-x}\,dx = \left. \left[ -x^2e^{-x}-2xe^{-x}-2e^{-x} \right] \right|_0^\infty = \bbox[red, 2pt]2$$

解答: $$\lim_{n\to \infty} \left| {x^{n+1} \over \ln(n+2)} \cdot {\ln(n+1)\over x^n}\right| =\lim_{n\to \infty} \left| {\ln(n+1) \over \ln(n+2)}x\right| \lt 1 \Rightarrow \bbox[red, 2pt]{|x|\lt 1}$$

解答: $$\textbf{(a)}\;  \lim_{x \to 0}{\int_0^{x^2}{t\over \sqrt{1+t^2}}dt\over x^4} = \lim_{x \to 0}{\left(\int_0^{x^2}{t\over \sqrt{1+t^2}}dt \right)'\over (x^4)'} = \lim_{x \to 0}{ {x^2\over \sqrt{1+x^4}}\cdot 2x\over 4x^3} = \lim_{x \to 0} {2\over 4\sqrt{1+x^4}} =\bbox[red, 2pt]{1\over 2} \\\textbf{(b)}\;  L =\left( e^x-1\right)^{1\over \ln x} \Rightarrow \ln L={\ln(e^x-1) \over  \ln x} \Rightarrow \lim_{x\to 0^+}\ln L=\lim_{x\to 0^+}{(\ln(e^x-1))' \over (  \ln x)'} =\lim_{x\to 0^+} {xe^x\over e^x-1} \\\qquad =\lim_{x\to 0^+} {(xe^x)'\over (e^x-1)'} =\lim_{x\to 0^+} {xe^x+e^x\over e^x}=\lim_{x\to 0^+} (x+1)=1 \Rightarrow \lim_{x\to 0^+} L=e^1= \bbox[red, 2pt] e \\ \textbf{(c)}\; L=\left(1+{1\over x} \right)^x \Rightarrow \ln L=x\ln \left(1+{1\over x} \right) \Rightarrow \lim_{x\to -\infty} \ln L=\lim_{x\to -\infty}{\ln\left(1+{1\over x} \right) \over {1\over x}} \\ =\lim_{x\to -\infty}{(\ln\left(1+{1\over x} \right))' \over ({1\over x})'} =\lim_{x\to -\infty}{  -{1\over x(x+1)} \over -{1\over x^2}} =\lim_{x\to -\infty} {x\over x+1} =1 \\ \Rightarrow \lim_{x\to -\infty} L=e^1=\bbox[red, 2pt]e \\\textbf{(d)}\; \lim_{n\to \infty} \left( {1\over n+1}+{1\over n+2}+ \cdots+ {1\over n+2n} \right) =\lim_{n\to \infty} \sum_{k=1}^n {1\over n+k}  =\lim_{n\to \infty} \sum_{k=1}^n {1/n\over 1+k/n} \\= \int_0^1 {1\over 1+x}\,dx =\bbox[red, 2pt]{\ln 2}$$

解答: $$\textbf{(a)}\; \text{By the triangle inequality, we have }||x|-|y|| \le |x-y|.\\ \text{ Now, }\lim_{x\to a} f(x)=L. \text{ That is, }\forall \epsilon\gt 0, \exists \delta \gt 0 \text{ such that } |x-a|\lt \delta \Rightarrow |f(x)-L|\lt \epsilon \\ \Rightarrow ||f(x)|-|L||\le |f(x)-L|\lt \epsilon \Rightarrow \lim_{x\to a}|f(x)|=|L|, \bbox[red, 2pt]{Q.E.D.} \\ \textbf{(b)}\; \text{An reverse example: Let }f(x)=\begin{cases} 2& x\ge a\\ -2 & x\lt a\end{cases}, \text{ then we have }\\\qquad \lim_{x\to a}|f(x)|=|2|, \text{ but }\lim_{x\to a}f(x) \text{ does not exist}$$

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