Find the value of $\mathop {\lim }\limits_{x \to \infty } \frac{{x + \cos x}}{{x – \cos x}}$.

$\begin{array}{cl} ◎\ &\because\ 1 \le \cos x \le 1 \\ & \therefore \frac{{ – 1}}{x} \le \frac{{\cos x}}{x} \le \frac{1}{x}\ (\text{for }x>0) \\ &\because \mathop {\lim }\limits_{x \to \infty } ( – \frac{1}{x}) = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0 \\&\therefore \mathop {\lim }\limits_{x \to \infty } \frac{{\cos x}}{x} = 0\ (\text{By sandwich lemma}) \end{array}$

$\mathop {\lim }\limits_{x \to \infty } \frac{{x + \cos x}}{{x – \cos x}}=\mathop {\lim }\limits_{x \to \infty } \frac{{1 + \frac{{\cos x}}{x}}}{{1 – \frac{{\cos x}}{x}}} = 1 ∎$

Find all horizontal asymptotes of graph of the function $f(x) = \frac{{\left|x \right|x}}{{x^2 + 1}}$.

$\mathop {\lim }\limits_{x \to \infty }\frac{{\left| x \right|x}}{{x^2 + 1}} = \mathop {\lim }\limits_{x \to \infty }\frac{{x^2 }}{{x^2 + 1}} = 1$

$\mathop {\lim }\limits_{x \to – \infty } \frac{{ – x^2 }}{{x^2 + 1}} = – 1$

horizontal asymptotes: $y=\pm 1 ∎$

Find the smallest positive $(x>0)$ inflection point of $F(x) = \int_0^x {\cos (t^{\frac{3}{2}} )} dt$.

$F'(x) = \cos (x^{\frac{3}{2}} ) \Rightarrow F^{\prime \prime} (x) = – \frac{3}{2}x^{\frac{1}{2}} \cdot \sin (x^{\frac{3}{2}} )\mathop = \limits^{{\rm{Let}}} 0$

$\Rightarrow x^{\frac{3}{2}} = k\pi$, $k=1,2,3,\dots$

$\Rightarrow$ the smallest positive inflection point appears at $x=\pi ^{\frac{2}{3}}$
$\Rightarrow$ the inflection point is $(\pi ^{\frac{2}{3}} ,\int_0^{\pi ^{\frac{2}{3}} } {\cos (t^{\frac{3}{2}} )} dt) ∎$

A building in the shape of a rectangular box is to have a volume of $12,000\ ft^3$. It is estimated that the annual heating and cooling costs will be \$2 / square foot for the top, \$ 4 / square foot for the front and back, and $3 / square foot for the sides. What is the minimal annual heating and cooling cost? 提示 有限制條件求極值時，使用拉格朗日乘數法。 解答 法一 Lagrange 乘數法 Let$f(x,y,z)=2xy+6yz+8xz$constraint:$xyz=12000$. Let$g(x,y,z)=xyz$Let$\nabla f+\lambda \nabla g=0\Rightarrow \left\{ {\begin{array}{c}
{2y + 8z + \lambda yz = 0} \\
{2x + 6z + \lambda xz = 0} \\
{6y + 8x + \lambda xy = 0} \\
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{c}
{2xy + 8xz + \lambda xyz = 0} \\
{2xy + 6yz + \lambda xyz = 0} \\
{6yz + 8xz + \lambda xyz = 0} \\
\end{array}} \right.\Rightarrow \left\{ {\begin{array}{c}
{8xz = 6yz \Rightarrow x:y = 3:4} \\
{2xy = 8xz \Rightarrow y:z = 4:1} \\
{2xy = 6yz \Rightarrow x:z = 3:1} \\ \end{array}} \right.\ \Rightarrow x:y:z = 3:4:1\Rightarrow\text{ Let} \left\{ {\begin{array}{c}{x = 3t} \\{y = 4t} \\{z = t} \\ \end{array}} \right.$,$t>0\because xyz=12000\therefore (3t)(4t)(t)=12000\Rightarrow 12t^3 = 12000 \Rightarrow t = 10 \Rightarrow \left\{ {\begin{array}{c}
{x = 30} \\
{y = 40} \\
{z = 10} \\
\end{array}} \right.\begin{array}{cl} \Rightarrow \text{minimum} & =f(30,40,10) \\ & =2400+2400+2400=7200　∎ \end{array}$法二 算幾不等式(簡答題或驗算時使用)$\begin{array}{cl} \frac{{{\rm{2}}xy{\rm{ + 6}}yz + 8xz}}{3} & \ge \sqrt[3]{{(2xy)(6yz)(8xz)}} \\ 2xz + 6yz + 8xz & \ge 3 \cdot \sqrt[3]{{2^2 \cdot 3 \cdot 2^3 \cdot 10^6 \cdot 4^2 \cdot 3^2 }} \\ & =3 \cdot 2 \cdot 10^2 \cdot 3 \cdot 4 = 7200　∎ \end{array}$講解影片YouTube課程平台 此題考點Lagrange 乘數法 第五題 (8 分) Find the values of$p$for which the function$f(x,y) = \left\{ {\begin{array}{c}{\frac{{(xy)^p }}{{x^4 + y^4 }},} &{{\rm{if }}(x,y) \ne (0,0)} \\ {0,} & {{\rm{if }}(x,y) = (0,0)} \\ \end{array}} \right.$is discontinuous at$(0,0)$. 提示 多變數函數尋找不連續點， 也就是計算極限，這題判斷是否收斂，需要妥善地處理分子跟分母的次方。 解答$\begin{array}{cl} \mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y)&=\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{(xy)^p }}{{x^4 + y^4 }} \\ & =\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{(xy)^{p – 2} (xy)^2 }}{{x^4 + y^4 }} \\ & =\mathop {\lim }\limits_{(x,y) \to (0,0)} (xy)^{p – 2} \frac{{(xy)^2 }}{{x^4 + y^4 }} \end{array}\begin{array}{cll} ◎\ & \because &(x^2 – y^2 )^2 = x^4 + y^4 – 2x^2 y^2 \\ &&\text{and }(x^2-y^2)\ge 0 \\ & \therefore & x^4+y^4\ge 2x^2y^2 \\&&\Rightarrow \frac{{x^2 y^2 }}{{x^4 + y^4 }} \le \frac{1}{2} \end{array}$So, if$p>2$,$\mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y) = \mathop {\lim }\limits_{(x,y) \to (0,0)} (xy)^{p – 2} \cdot \frac{{x^2 y^2 }}{{x^4 + y^4 }} = 0$i.e.$f(x,y)$is continuous at$(0,0)$. if$p\leq 2$,$\mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y) = \mathop {\lim }\limits_{(x,y) \to (0,0)} (xy)^{p – 2} \cdot \frac{{x^2 y^2 }}{{x^4 + y^4 }}$D.N.E. i.e.$f(x,y)$is discontinuous at$(0,0)$. Thus$f(x,y)$is discontinuous at$(0,0)$when$p\leq 2　∎$講解影片YouTube課程平台 此題考點二變數函數的極限 第六題 (8 分) Evaluate$\int_0^{\ln 5} {\int_{e^x }^5 {\frac{1}{{\ln y}}} } dydx$. 提示 遇到不好算的二重積分，可以嘗試極座標變換或交換積分順序 (Fubini 定理)。 解答$\begin{array}{cl} \int_0^{\ln 5} {\int_{e^x }^5 {\frac{1}{{\ln y}}} } dydx &=\int_1^5 {\int_0^{\ln y} {\frac{1}{{\ln y}}} } dxdy\\&=\int_1^5 {\frac{1}{{\ln y}}} \int_0^{\ln y} {dx} dy\\&= \int_1^5 1 dy=4　∎ \end{array}$講解影片YouTube課程平台 此題考點二變數函數的積分 第七題 (8 分) How many local extreme values does the function$f(x,y) = 10xye^{ – (x^2 + y^2 )}$have ? 提示 本題使用二次微分檢驗法。 解答$\begin{array}{cl} \nabla f =& (10ye^{ – (x^2 + y^2 )} + 10xye^{ – (x^2 + y^2 )} \cdot ( – 2x), \\ & 10xe^{ – (x^2 + y^2 )} + 10xye^{ – (x^2 + y^2 )} \cdot ( – 2y)) \end{array}\begin{array}{cl} \Rightarrow \nabla f =& (10ye^{ – (x^2 + y^2 )} \cdot (1 – 2x^2 ), \\ & 10xe^{ – (x^2 + y^2 )} \cdot (1 – 2y^2 )) \mathop = \limits^{{\rm{Let}}} (0,0) \end{array}\Rightarrow \left\{ {\begin{array}{c}
{y(1 – 2x^2 ) = 0 \Rightarrow y = 0{\text{ or }}x = \pm \frac{1}{{\sqrt 2 }}} \\
{x(1 – 2y^2 ) = 0 \Rightarrow x = 0{\text{ or }}y = \pm \frac{1}{{\sqrt 2 }}} \\
\end{array}} \right.\Rightarrow$critical points:$(0,0)$,$\pm (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$,$\pm (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})\begin{array}{cl} \text{H}f & =\left[ {\begin{array}{c}
{10ye^{ – (x^2 + y^2 )} \cdot ( – 2x)(1 – 2x^2 + 2)} & {(1 – 2y^2 )10e^{ – (x^2 + y^2 )} \cdot (1 – 2x^2 )} \\
{(1 – 2x^2 )10e^{ – (x^2 + y^2 )} \cdot (1 – 2y^2 )} & {10xe^{ – (x^2 + y^2 )} \cdot ( – 2y)(1 – 2y^2 + 2)} \\
\end{array}} \right] \\ & =10e^{ – (x^2 + y^2 )} \left[ {\begin{array}{c}
{ – 2xy(3 – 2x^2 )} & {(1 – 2y^2 )(1 – 2x^2 )} \\
{(1 – 2x^2 )(1 – 2y^2 )} & { – 2xy(3 – 2y^2 )} \\
\end{array}} \right]\\ \end{array}{\rm{D}}f(0,0) = {\rm{det}}(10\left[ {\begin{array}{c}
0 & 1 \\
1 & 0 \\
\end{array}} \right]) = – 100 < 0\Rightarrow (0,0,f(0,0))$is a saddle point.${\rm{D}}f( \pm (\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }})) = {\rm{det}}(10d^{ – 1} \left[ {\begin{array}{c}
{ – 2} & 0 \\
0 & { – 2} \\
\end{array}} \right]) = 40e^{ – 1} > 0\Rightarrow f( \pm (\frac{1}{{\sqrt 2 }}, – \frac{1}{{\sqrt 2 }}))$is a local extreme.${\rm{D}}f( \pm (\frac{1}{{\sqrt 2 }}, – \frac{1}{{\sqrt 2 }})) = {\rm{det}}(10e^{ – 1} \left[ {\begin{array}{c}
2 & 0 \\
0 & 2 \\
\end{array}} \right]) = 40e^{ – 1} > 0\Rightarrow f( \pm (\frac{1}{{\sqrt 2 }}, – \frac{1}{{\sqrt 2 }}))$is a local extreme.$\text{There are }4\text{ local extreme values.}　∎$講解影片YouTube課程平台 此題考點微分求極值 第八題 (8 分) Let$f(x,y) = kxye^{ – (x^2 + y^2 )}$be a joint probability density function on$D = \left\{ {0 < x < \infty ,0<y<\infty} \right\}$, then$k=?$提示 機率函數需要滿足其值介於 0 到 1 之間，及全部加總為 1。 解答$\iint_{D}f(x,y)dA=1\Rightarrow \int_0^\infty {\int_0^\infty {kxye^{ – (x^2 + y^2 )} } } dxdy = 1\Rightarrow \int_0^\infty {\int_0^\infty {xye^{ – x^2 } e^{ – y^2 } } } dxdy = \frac{1}{k}\Rightarrow \int_0^\infty {ye^{ – y^2 } } (\underbrace {\int_0^\infty {xe^{ – x^2 } } dx}_{\frac{1}{2}})dy = \frac{1}{2}\int_0^\infty {ye^{ – y^2 } } dy = \frac{1}{4}\Rightarrow k=4　∎$◎$\int_{\rm{0}}^\infty {xe^{ – x^2 } } dx=\int_0^\infty {e^{ – u} } \frac{{du}}{2}= \left[ { – \frac{1}{2}e^{ – u} } \right]_{u = 0}^{u = \infty }= \frac{1}{2}\left[ {0 – ( – 1)} \right] = \frac{1}{2}$講解影片YouTube課程平台 此題考點變數變換 乙.計算、證明題 第一題 (12 分) Goods 1 and 2 are available at prices (in dollars) of$p_1$per unit of good 1 and$p_2$per unit of good 2. A unility function$U(x_1,x_2)$is a function representing the unity or benefit for comsuming$x_j$units of goods$j$. The marginal unity of the$j$th good is$\frac{{\partial U}}{{\partial x_j }}$, the rate of increase in utility per unit increase in the$j$th good. Prove the following law of economics: Given a budget of$L$dollars, utility is maximized at the consumption level$(a,b)$where the ratio of marginal utility is equal to the ratio of prices:$\frac{{{\text{Marginal utility of good }}1}}{{{\text{Marginal utility of good }}2}} = \frac{{\partial U/\partial x_1 }}{{\partial U/\partial x_2 }} = \frac{{p_1 }}{{p_2 }}$提示 這題是限制條件的問題，要求利益函數的最大值，這時候要使用的就是 Lagrange 乘數法。 解答$U(x_1 ,x_2 ):$目標函數 constraint: Let$g(x_1 ,x_2 ) = x_1 p_1 + x_2 p_2 = L$Let$\nabla U + \lambda \nabla g = 0\Rightarrow (\frac{{\partial U}}{{\partial x_1 }},\frac{{\partial U}}{{\partial x_2 }}) + \lambda (p_1 ,p_2 ) = (0,0)\Rightarrow \left\{ {\begin{array}{c}
{\frac{{\partial U}}{{\partial x_1 }} + \lambda p_1 = 0} \\
{\frac{{\partial U}}{{\partial x_2 }} + \lambda p_2 = 0} \\
\end{array} \Rightarrow \left\{ {\begin{array}{c}
{\frac{{\partial U}}{{\partial x_1 }} = – \lambda p_1 \cdots (1) } \\
{\frac{{\partial U}}{{\partial x_2 }} = – \lambda p_2 \cdots (2) } \\
\end{array}} \right.} \right.\frac{(1)}{(2)}\Rightarrow \frac{{\partial U/\partial x_1 }}{{\partial U/\partial _2 }} = \frac{{ – \lambda p_1 }}{{ – \lambda p_2 }} = \frac{{p_1 }}{{p_2 }}　∎$講解影片YouTube課程平台 此題考點Lagrange 乘數法 第二題 (12 分) 〔1〕Determine whether the series$\sum\limits_{n = 1}^\infty {( – 1)^n \ln (1 + \frac{1}{n})}$diverges or converges conditionally or converges absolutely and give reasons for your answer. (6 分) 〔2〕Show that if$\sum\limits_{n = 1}^\infty {a_n }$converges, then$\sum\limits_{n = 1}^\infty {(\frac{{3 + \sin (a_n )}}{5})^n }$converges. (6 分) 提示 兩題為判別級數和收斂的基本題，八大審斂法與絕對、條件收斂需要很熟悉。 解答 〔1〕$\because \mathop {\lim }\limits_{n \to \infty } \ln (1 + \frac{1}{n}) = 0$and$\ln (1 + \frac{1}{n})$decreases.$\therefore \sum\limits_{n = 1}^\infty {( – 1)^n \ln (1 + \frac{1}{n})}$converges. (by alternating series)$\sum\limits_{n = 1}^\infty {\left| {( – 1)^n \ln (1 + \frac{1}{n})} \right|} = \sum\limits_{n = 1}^\infty {\ln (1 + \frac{1}{n})}\begin{array}{rl} \because & \mathop {\lim }\limits_{n \to \infty } \frac{{\ln (1 + \frac{1}{n})}}{{\frac{1}{n}}}=\mathop {\lim }\limits_{t \to 0^ + } \frac{{\ln (1 + t)}}{t}=\mathop = \limits^{\rm{L}} \mathop {\lim }\limits_{t \to 0^ + } \frac{{\frac{{\rm{1}}}{{{\rm{1 + }}t}}}}{{\rm{1}}} = 1\ (\text{let }t=\frac{1}{n}) \\ & \text{and } \sum\limits_{n = 1}^\infty {\frac{1}{n}}\text{ diverges.(by }p\text{-series test}) \\ \therefore &\sum\limits_{n = 1}^\infty {\ln (1 + \frac{1}{n})} \text{ diverges. (by limit comparison test)} 　∎\end{array}$附註：$\mathop = \limits^{\rm{L}}$表示此等號使用羅必達法則 (L’Hôpital’s rule) 〔2〕$\because \sum\limits_{n = 1}^\infty {a_n }$converges.$\therefore \mathop {\lim }\limits_{n \to \infty } a_n = 0\because \mathop {\lim }\limits_{n \to \infty } \left| {(\frac{{3 + \sin (a_n )}}{5})^n } \right|^{\frac{1}{n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{3 + \overbrace {\sin (a_n )}^0}}{5} = \frac{3}{5} < 1\therefore \sum\limits_{n = 1}^\infty {(\frac{{3 + \sin (a_n )}}{5})^n }$converges. (by root test)$　∎$講解影片YouTube課程平台 此題考點絕對收斂與條件收斂 第三題 (12 分) A trough with a trapezoidal cross section is to be constructed with a 1-foot base and sides that are 20 feet long and 1 foot wide, as shown in the figure. Only the angle$\theta$can be varied. What value of$\theta$will maximize the trough’s volume ? 提示 這題要求梯形容積最大的時機，因此只要把容積用$\theta$表示好，微分找最大值對應的$\theta$即可。 解答 area of trapezoidal$=\frac{{(2\sin \theta + 1) \cdot \cos \theta }}{2}= (1 + \sin \theta ) \cdot \cos \theta \mathop = \limits^{{\rm{Let}}} f(\theta )\begin{array}{cl} \Rightarrow f'(\theta) & =\cos \theta \cdot \cos \theta + (1 + \sin \theta ) \cdot ( – \sin \theta ) \\ & =\cos ^2 \theta – \sin ^2 \theta – \sin \theta \\ & =1 – 2\sin ^2 \theta – \sin \theta \mathop = \limits^{{\rm{Let}}} 0 \end{array}\Rightarrow 2\sin ^2 \theta + \sin \theta – 1 = (2\sin \theta – 1)(\sin \theta + 1) = 0\Rightarrow \sin \theta = \frac{1}{2}$or$\sin \theta=-1$(負不合$0^{\circ}\leq \theta \leq 90^{\circ}$) For$\sin \theta =\frac{1}{2}f(\theta ) = (1 + \frac{1}{2}) \times \frac{{\sqrt 3 }}{2} = \frac{{3\sqrt 3 }}{4}$For$\theta =0^{\circ}f(\theta)=1=\frac{4}{4}$When$\sin\theta =\frac{1}{2}$(i.e.$\theta =30^{\circ}$) the volume reaches its max.$　∎\$