Evaluate
〔1〕$\mathop {\lim }\limits_{x \to 3} \frac{{x^2 + x – 12}}{{x – 3}}$ (5分)
〔2〕$\mathop {\lim }\limits_{x \to 0} \frac{{x^2 }}{{\sec x – 1}}$. (5分)

〔1〕

$\mathop {\lim }\limits_{x \to 3} \frac{{x^2 + x – 12}}{{x – 3}}=\mathop {\lim }\limits_{x \to {\rm{3}}} \frac{{{\rm{(}}x{\rm{ + 4)(}}x – 3{\rm{)}}}}{{x – 3}} = 7 ∎$

$\mathop {\lim }\limits_{x \to 3} \frac{{x^2 + x – 12}}{{x – 3}}\mathop {\rm{ = }}\limits^{\rm{L}} \mathop {\lim }\limits_{x \to {\rm{3}}} \frac{{{\rm{2}}x{\rm{ + 1}}}}{{\rm{1}}} = 7 ∎$

〔2〕

$\begin{array}{cl} \mathop {\lim }\limits_{x \to 0} \frac{{x^2 }}{{\sec x – 1}} & =\mathop {\lim }\limits_{x \to {\rm{0}}} \frac{{x^2 }}{{\sec x – 1}} \cdot \frac{{\sec x + 1}}{{\sec x + 1}} \\ & =\mathop {\lim }\limits_{x \to 0} \frac{{x^2 (\sec x + 1)}}{{\sec ^2 x – 1}} \\ & =\mathop {\lim }\limits_{x \to 0} \frac{{x^2 (\sec x + 1)}}{{\tan ^2 x}} \\&=\mathop {\lim }\limits_{x \to 0} \underbrace {(\frac{{x^2 }}{{\sin ^2 x}})}_{\to{\rm{ 1}}} \cdot \underbrace {\cos ^2 x}_{\to{\rm{ 1}}} \cdot \underbrace {(\sec x + 1)}_{\to{\rm{ 2}}} = 2 ∎ \end{array}$

$\begin{array}{cl} \mathop {\lim }\limits_{x \to 0} \frac{{x^2 }}{{\sec x – 1}} & =\mathop {\lim }\limits_{x \to {\rm{0}}} \frac{{{\rm{2}}x}}{{\sec x\tan x}} \\ & =\mathop {\lim }\limits_{x \to 0} \underbrace {(\frac{x}{{\sin x}})}_{\to{\rm{ 1}}} \cdot \underbrace {(\frac{2}{{\sec x}})}_{\to{\rm{ 2}}} \cdot \underbrace {\cos x}_{\to{\rm{ }}1} = 2 ∎ \end{array}$

Evaluate $\left. {\frac{{\partial f}}{{\partial x}}} \right|_{(0,0)}$ and $\left. {\frac{{\partial f}}{{\partial y}}} \right|_{(0,0)}$ for $f(x,y) = \left\{ {\begin{array}{cl}{\frac{{3x^4 + xy^2 }}{{2x^3 + 4xy + y}},} & {(x,y) \ne (0,0)} \\{0,} & {(x,y) = (0,0)} \end{array}} \right.$

$\begin{array}{cl} \left. {\frac{{\partial f}}{{\partial x}}} \right|_{(0,0)} & =\mathop {\lim }\limits_{h \to {\rm{0}}} \frac{{f(h,0) – f(0,0)}}{h} \\ & =\mathop {\lim }\limits_{h \to {\rm{0}}} \frac{{\frac{{3h^4 }}{{2h^3 }} – 0}}{h} = \frac{3}{2} ∎ \end{array}$

$\begin{array}{cl} \left. {\frac{{\partial f}}{{\partial y}}} \right|_{(0,0)}&=\mathop {\lim }\limits_{k \to 0} \frac{{f(0,k) – f(0,0)}}{k} \\ & =\mathop {\lim }\limits_{k \to 0} \frac{{\frac{0}{k} – 0}}{k} = 0 ∎\end{array}$

Given $f(t) = \left\{ {\begin{array}{cl}{1,} & {t \le 0} \\ {1 – t,} & {t > 0} \\ \end{array}} \right.$ and $F(x) = \int_{ – 1}^{2ax + 2} {f(t)} dt$ with $a>0$, find $a$ so that $F$ is maximum at $x=-2a$.

$F'(x){\rm{ = }}f(2ax + 2) \cdot 2a$
$\because F$ has max at $x=-2a$
$\therefore F'(-2a)=0$
$f( – 4a^2 + 2) \cdot 2a = 0$
$\Rightarrow f(\underbrace { – 4a^2 + 2}_1) = 0$
$\Rightarrow – 4a^2 + 2 = 1$
$\Rightarrow a-\pm \frac{1}{2}\ \because a>0\ \therefore a=\frac{1}{2} ∎$

Find the largest possible area of a triangle with vertices $(0,2)$, $(1,0)$ and the third vertex on the ellipse $x^2 + \frac{{y^2 }}{4} = 1$.

$x^2+\frac{y^2}{4}=1$
$\Rightarrow 2x+\frac{1}{4}\cdot 2yy’=0$ (切線斜率和點的關係式)

Let $y’=-2$
$\Rightarrow 2x+\frac{1}{2} y(-2)=0$
$\Rightarrow 2x-y=0$

$P: \left\{ \begin{array}{l} 2x-y=0\Rightarrow y=2x \\ x^2+\frac{y^2}{4}=1 \end{array}\right.$
$\Rightarrow x^2 + \frac{{4x^2 }}{4} = 1$
$\Rightarrow x = \pm \frac{1}{{\sqrt 2 }}$
$\Rightarrow (x,y) = \pm (\frac{1}{{\sqrt 2 }},\sqrt 2 )$

$\Rightarrow$ 所求$=\frac{1}{2}\sqrt {1^2 + 2^2 } \cdot \frac{{\left| { – \sqrt 2 – \sqrt 2 – 2} \right|}}{{\sqrt {2^2 + 1^2 } }} = 1 + \sqrt 2$

Evaluate $\int_{\ln \frac{1}{4}}^{\ln \frac{1}{2}} {\frac{{e^x }}{{\sqrt {1 – 4e^{2x} } }}} dx$.

Let $u=e^x$ ($du=e^x dx$)
$\int_{\ln \frac{1}{4}}^{\ln \frac{1}{2}} {\frac{{e^x }}{{\sqrt {1 – 4e^{2x} } }}} dx=\int_{\frac{1}{4}}^{\frac{1}{2}} {\frac{1}{{\sqrt {1 – 4u^2 } }}} du = \int_{\frac{1}{4}}^{\frac{1}{2}} {\frac{1}{{\sqrt {1^2 – (2u)^2 } }}} du$

Let $2u=\sin v$ ($2du=\cos vdv$)
$\begin{array}{cl} \int_{\frac{1}{4}}^{\frac{1}{2}} {\frac{1}{{\sqrt {1^2 – (2u)^2 } }}} du & =\int_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{1}{{\cos v}} \cdot \frac{{\cos v}}{2}} dv \\ & =\frac{1}{2}\int_{\frac{\pi }{6}}^{\frac{\pi }{2}} {dv} \\ & =\frac{1}{2}(\frac{\pi }{2} – \frac{\pi }{6}) = \frac{\pi }{6} ∎ \end{array}$

Evaluate $\int_0^{\frac{{\sqrt 2 }}{2}} {\int_y^{\sqrt {1 – y^2 } } {e^{x^2 + y^2 } } } dxdy$.

Use polar coordinate:
Let $\left\{ \begin{array}{c}{x = r\cos \theta } \\{y = r\sin \theta } \end{array}\right.\Rightarrow \left\{ \begin{array}{l} r^2=x^2+y^2\\ dxdy=rdrd\theta \end{array}\right.$

$\int_0^{\frac{{\sqrt 2 }}{2}} {\int_y^{\sqrt {1 – y^2 } } {e^{x^2 + y^2 } } } dxdy=\int_0^{\frac{\pi }{4}} {\int_0^1 {e^{r^2 } } } rdrd\theta$

Let $u=r^2$ ($du=2rdr$)
$\begin{array}{cl} \int_0^{\frac{\pi }{4}} {\int_0^1 {e^{r^2 } } } rdrd\theta &= \int_0^{\frac{\pi }{4}} {\int_0^1 {e^u \frac{{du}}{2}} } d\theta \\ &=\frac{1}{2}\int_0^{\frac{\pi }{4}} {\left[ {e^u } \right]_{u = 0}^{u = 1} } d\theta \\ &= \frac{1}{2}\int_0^{\frac{\pi }{4}} {(e – 1)} d\theta\\&= \frac{1}{2}(e – 1)\frac{\pi }{4} = \frac{\pi }{8}(e – 1) ∎ \end{array}$

Derive the complete Taylor series expansion for $\ln (\frac{{1 + 2x}}{{1 – 2x}})$ about $x=0$. (In the form $\sum\limits_{k = 1}^\infty {a_k x^{2k – 1} }$ with a general formula for $a_k$.)

$\begin{array}{cl} ◎\ \ln \left| {1 + x} \right| = \int_0^x {\frac{1}{{1 + t}}} dt& =\int_0^x {(1 – t + t^2 – t^3 + \cdots )} dt,\ \left| t \right| < 1\\ & =x – \frac{{x^2 }}{2} + \frac{{x^3 }}{3} – \frac{{x^4 }}{4} + \cdots ,\ \left| x \right| < 1 \end{array}$

$\ln (1 + 2x)=2x – \frac{{2^2 x^2 }}{2} + \frac{{2^3 }}{3} – \frac{{2^4 x^4 }}{4} + \frac{{2^5 x^5 }}{5} – \frac{{2^6 x^6 }}{6} + \cdots$, $-\frac{1}{2} < x < \frac{1}{2}$
$\ln (1 – 2x)=-2x – \frac{{2^2 x^2 }}{2} – \frac{{2^3 }}{3} – \frac{{2^4 x^4 }}{4} – \frac{{2^5 x^5 }}{5} – \frac{{2^6 x^6 }}{6} – \cdots$, $-\frac{1}{2} < x < \frac{1}{2}$

$\begin{array}{cl} \ln (\frac{{1 + 2x}}{{1 – 2x}}) &=\ln(1+2x)-\ln(1-2x)\\& =2(2x + \frac{{2^3 x^3 }}{3} + \frac{{2^5 x^5 }}{5} + \cdots ) \\ & =\underbrace {2^2 }_{a_1 }x +\underbrace {\frac{{2^4 }}{3}}_{a_2 }x^3 +\underbrace {\frac{{2^5 }}{5}}_{a_3 }x^5 + \cdots \end{array}$
$\Rightarrow a_k = \frac{{2^{2k} }}{{2k – 1}},\ (k=0,1,2,\cdots) ∎$

Given function $T(x,y)=1+x^2-y^2$, find the curve $\gamma (t)=(x(t),y(t))$ so that $\gamma (0)=(1,4)$ and $\gamma ‘(t)=-\nabla T(\gamma (t))$.

$\because \gamma (0) = (x(0),y(0)) = (1,4)$
$\therefore \left\{ {\begin{array}{c} {x(0) = 1} \\ {y(0) = 4} \\ \end{array}} \right.$

$\nabla T = (2x, – 2y)$
$\Rightarrow – \nabla T = ( – 2x,2y)$
$\begin{array}{cl} \Rightarrow – \nabla T(\gamma (t)) & =\nabla T(x(t),y(t)) \\ & =( – 2x(t),2y(t)) \end{array}$
$\because \gamma ^{\prime}(t) = – \nabla T(\gamma (t))$
$\therefore (x'(t),y'(t)) = ( – 2x(t),2y(t))$
$\Rightarrow \left\{ {\begin{array}{l} {x'(t) = – 2x(t)} \\ {y'(t) = 2y(t) } \\ \end{array}} \right.$

$\because \left\{ \begin{array}{l}x'(t) = – 2x(t) \\x(0) = 1 \end{array} \right.$ and $\left\{ \begin{array}{l} y'(t) = 2y(t)\\ y(0) = 4 \end{array} \right.$
$\begin{array}{cl} \therefore & x(t) = 1 \cdot e^{ – 2t} = e^{ – 2t} \\ & y(t) = 4e^{2t} \end{array}$
$\Rightarrow \gamma (t) = (e^{ – 2t} ,4e^{2t} ) ∎$

Find $(a,b)$ with $\frac{1}{2} \le b \le \frac{1}{2}$ so that
● The point $P=(1,a,b)$ is on the surface $E$ defined by $\frac{x}{2} – \frac{y}{4} + \frac{{\sin (2z)}}{4} = 0$
● The tangent plane to $E$ at $P$ contains lines
$\ell _1 (t) = P + t(\frac{1}{2},1,0)$ and $\ell_2 (t)=P+t(0,2,1)$

$\overset{\rightharpoonup}{n} \mathbin{/\mkern-5mu/}\nabla E = (\frac{1}{2}, – \frac{1}{4},\frac{{\cos (2b)}}{2})$

$(1,2,0) \times (0,2,1) = (2, – 1,2) \Rightarrow \overset{\rightharpoonup}{n} \mathbin{/\mkern-5mu/}(2,-1,2)$

$\Rightarrow \cos (2b)=1 \Rightarrow b=k\pi ,\ k\in \mathbb{Z}$
$\because \frac{1}{2} \le b \le \frac{1}{2}$
$\therefore b=0$
$\Rightarrow P=(1,a,0)$

$\because P\in E:\ \frac{x}{2} – \frac{y}{4} + \frac{{\sin (2z)}}{4} = 0$
$\therefore \frac{1}{2} – \frac{a}{4} + \frac{{\sin 0}}{4} = 0$
$\Rightarrow a = 2 \Rightarrow P(1,2,0) \Rightarrow (a,b) = (2,0) ∎$

Evaluate $\int_{L}\textbf{F}\cdot d\textbf{r}$, where $\textbf{F}=(4x + 5y,e^{\cos y} + 7x)$, and $L$ is the path from $\textbf{A}$ to $\textbf{B}$, to $\textbf{C}$ to $\textbf{A}$ to $\textbf{D}$ as shown below:

Note: The path from $\textbf{B}$ to $\textbf{C}$ is circular.

Let $S$ be the path from $\textbf{A}$ to $\textbf{B}$ to $\textbf{C}$ to $\textbf{A}$
$\begin{array}{cl} \oint_S \textbf{F}\cdot d\textbf{r} & =\oint_S {(4x + 5y,e^{\cos y} + 7x) \cdot (dx,dy)} \\ & =\oint_S {\underbrace {(4x + 5y)}_Pdx + \underbrace {(e^{\cos y} + 7x)}_Qdy} \\ & =\iint_{\Omega}(e^{\cos y + 7x} )_x – (4x + 5y)_y dA \\&=\iint_{\Omega}(7-5)dA=2\cdot \frac{\pi}{4}=\frac{\pi}{2} \end{array}$

Let $L’$ be the path from $\textbf{A}$ to $\textbf{D}$.
$L’:\ (x,y)=(t,0)$, $t$ from $0$ to $-1$
$\begin{array}{cl} \Rightarrow \oint_{L’} \textbf{F}\cdot d\textbf{r} & =\oint_S {(4x + 5y,e^{\cos y} + 7x) \cdot (dx,dy)} \\ & =\int_0^{ – 1} {(4t,e + 7t) \cdot (dt,0)} \\ & =\int_0^{ – 1} {4tdt} = \left. {2t^2 } \right|_0^{ – 1} = 2 \end{array}$

$\Rightarrow \int_L \textbf{F} \cdot d\textbf{r} = \frac{\pi }{2} + 2 ∎$