112年中山大學碩士班－微積分詳解

 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： $${d\over dx}\left({\ln x\over e^{\sin x}} \right) ={1\over xe^{\sin x}}-{\ln x\over (e^{\sin x})^2}\cdot \cos x\cdot e^{\sin x} =\bbox[red, 2pt]{{1\over xe^{\sin x}}-{\cos x\ln x\over e^{\sin x}}}$$

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： $$a_{n}={(x-2)^n \over n+2} \Rightarrow \left| {a_{n+1}\over a_n}\right| = \left| {(x-2)^{n+1}\over n+3} \cdot {n+2\over (x-2)^n}\right| = \left| {n+2\over n+3}\cdot (x-2)\right| \\ 因此\lim_{n\to \infty}\left| {a_{n+1}\over a_n}\right| =|x-2|\lt 1 \Rightarrow 1\lt x\lt 3 \\ x=1 \Rightarrow \sum_{n=0}^\infty {(x-2)^n\over n+2}= \sum_{n=0}^\infty {(-1)^n\over n+2} ={1\over 2}-{1\over 3}+{1\over 4}-\cdots =1-\ln 2\\ x=3 \Rightarrow \sum_{n=0}^\infty {(x-2)^n\over n+2}= \sum_{n=0}^ \infty {1\over n+2} 為調和級數＝\infty\\ 因此 收斂區間：\bbox[red,2pt]{[1,3)}$$

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： $$\mathbf{(i)} \text{range of }y:\bbox[red, 2pt]{(-{\pi\over 2},{\pi\over 2})}, \text{domain of }y: \bbox[red,2pt]{(-\infty, \infty)}\\ \mathbf{(ii)}\; y=\arctan x \Rightarrow \tan y=x \Rightarrow{d\over dx} \tan y= {d\over dx}x \Rightarrow \sec^2 y\cdot {dy\over dx}=1,\bbox[red, 2pt]{故得證}\\ \tan y=x \Rightarrow \sec y={1\over \sqrt{1+x^2}} \Rightarrow {d\over dx}f^{-1}(x)={dy\over dx} ={1\over \sec^2(y)}=\bbox[red, 2pt]{1\over 1+x^2}$$

解答： $$\mathbf{(i)}\;{d\over dx}\sinh^{-1}={1\over \sqrt{1+x^2}}  \Rightarrow \int{1\over \sqrt{x^2+2x+2}}dx = \int {1\over (x+1)^2+1}dx = \bbox[red, 2pt]{\sinh^{-1}(x+1)+C} \\ \mathbf{(ii)}\; \int {x\over (x+1)(x+2)^2}dx = \int {-1\over x+1}+{1\over x+2}+ {2\over (x+2)^2}\,dx\\= \bbox[red, 2pt]{-\ln(x+1)+\ln(x+2)-{2\over x+2}+C} \\\mathbf{(iii)}\; 令\cases{u=x \\ dv=\sin xdx} \Rightarrow \cases{du=dx\\ v=-\cos x} \Rightarrow \int x\sin x\,dx =-x\cos x+ \int \cos x\,dx\\= \sin x-x\cos x+C \Rightarrow \int_0^{\pi/2} x\sin x\,dx = \left. \left[  \sin x-x\cos x\right]\right|_0^{\pi/2} =\bbox[red, 2pt]1$$