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:$$\textbf{(a)}\; \lim_{x \to 0}{\int_0^{x^2}{t\over \sqrt{1+t^2}}dt\over x^4} = \lim_{x \to 0}{\left(\int_0^{x^2}{t\over \sqrt{1+t^2}}dt \right)'\over (x^4)'} = \lim_{x \to 0}{ {x^2\over \sqrt{1+x^4}}\cdot 2x\over 4x^3} = \lim_{x \to 0} {2\over 4\sqrt{1+x^4}} =\bbox[red, 2pt]{1\over 2} \\\textbf{(b)}\; L =\left( e^x-1\right)^{1\over \ln x} \Rightarrow \ln L={\ln(e^x-1) \over \ln x} \Rightarrow \lim_{x\to 0^+}\ln L=\lim_{x\to 0^+}{(\ln(e^x-1))' \over ( \ln x)'} =\lim_{x\to 0^+} {xe^x\over e^x-1} \\\qquad =\lim_{x\to 0^+} {(xe^x)'\over (e^x-1)'} =\lim_{x\to 0^+} {xe^x+e^x\over e^x}=\lim_{x\to 0^+} (x+1)=1 \Rightarrow \lim_{x\to 0^+} L=e^1= \bbox[red, 2pt] e \\ \textbf{(c)}\; L=\left(1+{1\over x} \right)^x \Rightarrow \ln L=x\ln \left(1+{1\over x} \right) \Rightarrow \lim_{x\to -\infty} \ln L=\lim_{x\to -\infty}{\ln\left(1+{1\over x} \right) \over {1\over x}} \\ =\lim_{x\to -\infty}{(\ln\left(1+{1\over x} \right))' \over ({1\over x})'} =\lim_{x\to -\infty}{ -{1\over x(x+1)} \over -{1\over x^2}} =\lim_{x\to -\infty} {x\over x+1} =1 \\ \Rightarrow \lim_{x\to -\infty} L=e^1=\bbox[red, 2pt]e \\\textbf{(d)}\; \lim_{n\to \infty} \left( {1\over n+1}+{1\over n+2}+ \cdots+ {1\over n+2n} \right) =\lim_{n\to \infty} \sum_{k=1}^n {1\over n+k} =\lim_{n\to \infty} \sum_{k=1}^n {1/n\over 1+k/n} \\= \int_0^1 {1\over 1+x}\,dx =\bbox[red, 2pt]{\ln 2}$$
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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解題僅供參考,其他歷年試題及詳解
第三題(a),x^4羅畢達後錯了是4x^3,所以極限不對
回覆刪除已修訂,謝謝
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