國立中山大學112學年度碩士班暨碩士在職專班
科目名稱:微積分【企管系企管甲班碩士班甲組選考,乙組選考,丙組選考】
Q1. (5+5=10%)
Calculate the following limits:
(i)
解答:$$-{1\over x}\le {\sin x\over x}\le {1\over x} \Rightarrow \lim_{x \rightarrow \infty}-{1\over x}\le \lim_{x \rightarrow \infty}{\sin x\over x} \le \lim_{x \rightarrow \infty}{1\over x} \Rightarrow 0\le \lim_{x \rightarrow \infty}{\sin x\over x} \le 0\\ 依夾擠定理: \lim_{x \rightarrow \infty}{\sin x\over x} =\bbox[red, 2pt]0$$
(ii)解答:$$\lim_{x \rightarrow 3}{5x^2-8x-13\over x^2-5} ={45-24-13\over 9-5} =\bbox[red, 2pt]2$$
Q2. (10%)Evaluate the following differentiation:
解答:$$(x-y)^2=x+y-1 \Rightarrow 2(x-y)(1-y')=1+y' \Rightarrow {dy\over dx}=y'=\bbox[red,2pt]{{2(x-y) -1\over 2(x-y)+1}}$$
解答:$$a_{n}={(x-2)^n \over n+2} \Rightarrow \left| {a_{n+1}\over a_n}\right| = \left| {(x-2)^{n+1}\over n+3} \cdot {n+2\over (x-2)^n}\right| = \left| {n+2\over n+3}\cdot (x-2)\right| \\ 因此\lim_{n\to \infty}\left| {a_{n+1}\over a_n}\right| =|x-2|\lt 1 \Rightarrow 1\lt x\lt 3 \\ x=1 \Rightarrow \sum_{n=0}^\infty {(x-2)^n\over n+2}= \sum_{n=0}^\infty {(-1)^n\over n+2} ={1\over 2}-{1\over 3}+{1\over 4}-\cdots =1-\ln 2\\ x=3 \Rightarrow \sum_{n=0}^\infty {(x-2)^n\over n+2}= \sum_{n=0}^
\infty {1\over n+2} 為調和級數=\infty\\ 因此 收斂區間:\bbox[red,2pt]{[1,3)}$$
\infty {1\over n+2} 為調和級數=\infty\\ 因此 收斂區間:\bbox[red,2pt]{[1,3)}$$
解答:$$\ln(x+2) =\int{1\over x+2}\,dx ={1\over 2}\int {1\over 1-(-x/2)}dx \\ ={1\over 2}\int (1-{x\over 2}+{x^2\over 4}-{x^3\over 8}+\cdots )\,dx ={1\over 2}(x-{1\over 4}x^2+{1\over 12}x^3-{1\over 32}x^4+\cdots)\\ \Rightarrow f(x)=x\ln(x+2)= \bbox[red, 2pt]{{1\over 2}x^2-{1\over 8}x^3}+{1\over 24}x^4-{1\over 64}x^5+\cdots $$
解答:$$\mathbf{(i)} \text{range of }y:\bbox[red, 2pt]{(-{\pi\over 2},{\pi\over 2})}, \text{domain of }y: \bbox[red,2pt]{(-\infty, \infty)}\\ \mathbf{(ii)}\; y=\arctan x \Rightarrow \tan y=x \Rightarrow{d\over dx} \tan y= {d\over dx}x \Rightarrow \sec^2 y\cdot {dy\over dx}=1,\bbox[red, 2pt]{故得證}\\ \tan y=x \Rightarrow \sec y={1\over \sqrt{1+x^2}} \Rightarrow {d\over dx}f^{-1}(x)={dy\over dx} ={1\over \sec^2(y)}=\bbox[red, 2pt]{1\over 1+x^2}$$
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