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

〔1〕
$\mathop {\lim }\limits_{n \to \infty } \frac{n}{{n + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{1 + \frac{1}{n}}} = 1 ∎$

〔2〕
$\begin{array}{cl} \mathop {\lim }\limits_{x \to 0} e^{ – \frac{1}{{x^2 }}} \cdot \frac{{1 – 3x^2 }}{{x^5 }} & =\mathop {\lim }\limits_{x \to 0} \frac{{e^{ – \frac{1}{{x^2 }}} }}{{x^5 }}(1 – 3x^2 ) \\ & =\mathop {\lim }\limits_{x \to 0} \frac{{e^{ – \frac{1}{{x^2 }}} }}{{x^6 }}(x – 3x^3 ) \end{array}$

Let $t=\frac{1}{x^2}$
$\because\ \mathop {\lim }\limits_{x \to 0} \frac{{e^{ – \frac{1}{{x^2 }}} }}{{x^6 }} = \mathop {\lim }\limits_{t \to \infty } t^3 e^{ – t} = \mathop {\lim }\limits_{t \to \infty } \frac{{t^3 }}{{e^t }} = 0$
$\therefore\$ 所求$=\left[ {\mathop {\lim }\limits_{x \to 0} \frac{{e^{ – \frac{1}{{x^2 }}} }}{{x^6 }}} \right] \cdot \left[ {\mathop {\lim }\limits_{x \to 0} (x – 3x^3 )} \right] = 0 \cdot 0 ∎$

Find the point on the curve $y=x^{\frac{3}{2}}$ that is closed to the point $(\frac{5}{2},0)$.

$\Rightarrow d=\sqrt{(t – \frac{5}{2})^2 + t^3}$
Let $f(t) = (t – \frac{5}{2})^2 + t^3$
$\Rightarrow f'(t) = 2(t – \frac{5}{2}) + 3t^2 \mathop = \limits^{{\text{Let}}} 0$
$\Rightarrow 2t – 5 + 3t^2 = 0 \Rightarrow (3t + 5)(t – 1) = 0$
$t=-\frac{5}{3}$(不合) or $t=1$

Find the length of the parametric curve $x=3t^2,\ y=2t^3,\ 0\leq t\leq 1$.

$\begin{array}{cl} \ell & =\int_0^1 {\sqrt {(6t)^2 + (6t^2 )^2 } } dt \\ & =\int_0^1 {\sqrt {36t^2 + 36t^4 } } dt \\ & =\int_0^1 {6t\sqrt {1 + t^2 } } dt\\&(\text{Let }u=1+t^2\Rightarrow du=2tdt)\\&=\int_0^1 {\sqrt u \cdot 3} du\\&=3 \cdot \left. {\frac{{u^{\frac{3}{2}} }}{{\frac{3}{2}}}} \right|_{u = 1}^{u = 2}\\&=2(2^{\frac{3}{2}} – 1) = 4\sqrt 2 – 2 ∎ \end{array}$

Determine whether the improper integral $\int_1^\infty {\frac{{1 – \cos x}}{{x^2 }}} dx$ is convergent or divergent ?

Note that $0 \le \int_1^\infty {\frac{{1 – \cos x}}{{x^2 }}} dx \le \int_1^\infty {\frac{2}{{x^2 }}} dx$

Check $\int_1^\infty {\frac{2}{{x^2 }}dx}$ is convergent.
$\because\ \sum\limits_{n = 1}^\infty {\frac{2}{{n^2 }}}$ (by $p$-series)
$\therefore\ \int_1^\infty {\frac{2}{{x^2 }}} dx$ converges (by integral test).

$\Rightarrow \int_1^\infty {\frac{{1 – \cos x}}{{x^2 }}} dx$ converges (by comparison test). $∎$

Find the radius of convergence of the power series $\sum\limits_{n = 1}^\infty {n^2 x^n }$ and evaluate $\sum\limits_{n = 1}^\infty {\frac{{n^2 }}{{2^n }}}$.

radius$=\frac{{\rm{1}}}{{\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{(n + 1)^2 }}{{n^2 }}} \right|}}{\rm{ = }}\frac{1}{1} = 1 ∎$

Let $S = \sum\limits_{n = 1}^\infty {\frac{{n^2 }}{{2^n }}}$
$S = \frac{{1^2 }}{{2^1 }} + \frac{{2^2 }}{{2^2 }} + \frac{{3^2 }}{{2^3 }} + \frac{{4^2 }}{{2^4 }} + \cdots$
$\frac{1}{2}S=\frac{{1^2 }}{{2^2 }} + \frac{{2^2 }}{{2^3 }} + \frac{{3^2 }}{{2^4 }} + \cdots$
$S-\frac{1}{2}S=\frac{1}{2}S=\frac{1}{2} + \frac{3}{{2^2 }} + \frac{5}{{2^3 }} + \frac{7}{{2^4 }} + \cdots$
$\Rightarrow \frac{1}{4}S=\frac{1}{{2^2 }} + \frac{3}{{2^3 }} + \frac{5}{{2^4 }} + \cdots$
$\frac{1}{2}S-\frac{1}{4}S=\frac{1}{4}S=\frac{1}{2} + \frac{2}{{2^2 }} + \frac{2}{{2^3 }} + \frac{2}{{2^4 }} + \cdots$
$\Rightarrow \frac{1}{4}S = \frac{1}{2} + \frac{{\frac{1}{2}}}{{1 – \frac{1}{2}}} = \frac{3}{2}$
$\Rightarrow S = \sum\limits_{n = 1}^\infty {\frac{{n^2 }}{{2^n }}} = \frac{3}{2} \times 4 = 6 ∎$

Find the equation of the tangent plane and the normal line to the surface $S:x^2 y + e^{xyz} – 2\cos (xz) = 0$ at $(1,1,0)$.

Let $f(x,y,z){\rm{ = }}x^2 y + e^{xyz} – 2\cos (xz)$
$\nabla f(x,y,z) = (2xy + yze^{xyz} + 2z\sin (xz),x^2 + xze^{xyz} ,xye^{xyz} + 2x\sin (xz))$
$\Rightarrow \overset{\rightharpoonup}{n}= \nabla f(1,1,0) = (2,1,1)$

tangent plane: $2x+y+z=3$
normal line: $\left\{ \begin{array}{l}x=1+2t\\ y=1+t \\ z=t \end{array}\right. ,\ t\in \mathbb{R} ∎$

Let $f:\mathbb{R}^2\to\mathbb{R}$ be a function. Assume that all second partial derivative of $f$ exist and continuous and $f_{xx}+f_{yy}=0$, for all $(x,y)\in\mathbb{R}^2$.
Define $g:\ \mathbb{R}^2\to\mathbb{R}$ by $g(u,v) = f(u^2 – v^2 ,2uv)$. Find $g_{uu}+g_{vv}$.

$\left\{ \begin{array}{l} x = u^2 – v^2 \\ y = 2uv \end{array} \right. \Rightarrow \left\{ \begin{array}{cc} x_u = 2u, & x_v = – 2v \\ y_u = 2v, & y_v = 2u \end{array} \right.$
$g_u = f_x \cdot x_u + f_y \cdot y_u = f_x \cdot (2u) + f_y \cdot (2v)$
$\begin{array}{rl} \Rightarrow g_{uu}= & (f_{xx} \cdot x_u + f_{xy} \cdot y_u ) \cdot (2u) + f_x \cdot (2)\\& + (f_{yx} \cdot x_u + f_{yy} \cdot y_u ) \cdot (2v) + f_y \cdot (0) \\ =&f_{xx} \cdot (4u^2 ) + f_{xy} \cdot (4uv) + 2f_x + f_{yx} \cdot (4uv) + f_{yy} \cdot (4v^2 ) \end{array}$

$g_v = f_x \cdot x_v + f_y \cdot y_v = f_x \cdot ( – 2v) + f_y \cdot (2u)$
$\begin{array}{rl} \Rightarrow g_{vv} = & (f_{xx} \cdot x_v + f_{xy} \cdot y_v ) \cdot ( – 2v) + f_x \cdot ( – 2) \\ & +(f_{yx} \cdot x_v + f_{yy} \cdot y_v ) \cdot (2u) + f_y \cdot (0) \\ =& f_{xx} \cdot (4v^2 ) + f_{xy} \cdot ( – 4uv) + f_x \cdot ( – 2) + f_{yx} \cdot ( – 4uv) + f_{yy} \cdot (4u^2 ) \end{array}$

$\begin{array}{cl} \Rightarrow g_{uu} + g_{vv} & =f_{xx} \cdot (4u^2 + 4v^2 ) + f_{yy} \cdot (4u^2 + 4v^2 ) \\ & =(4u^2 + 4v^2 ) \cdot \underbrace {(f_{xx} + f_{yy} )}_0 = 0 ∎ \end{array}$

Evaluate the double integral $\iint_{D}\sin (x+y)dA$ where $D$ is the region bounded by $x+y=\pi$, $x+y=0$ and $x-y=\pi$ and $x-y=0$.

Let $\left\{ {\begin{array}{l} {u = x + y} \\ {v = x – y}\end{array} \Rightarrow \frac{{dudv}}{{dxdy}} = \left| {\begin{array}{c} 1 & 1 \\ 1 & – 1 \end{array}} \right| = – 2} \right.$

$\begin{array}{cl} \iint_{D}\sin (x+y)dxdy & =\iint_{D’}\sin u \cdot \underbrace {\left| {\frac{{dxdy}}{{dudv}}} \right|}_{\frac{1}{2}}dudv \\ & =\frac{1}{2}\int_0^\pi {\int_0^\pi {\sin u} } dudv \\ & =\frac{1}{2}\int_0^\pi {\left. { – \cos u} \right|_{u = 0}^{u = \pi } } dv\\&=\frac{1}{2}\int_0^\pi {1 – ( – 1)} dv = \pi ∎ \end{array}$

Evaluate the triple integral $\iiint_E\sqrt{x^2+y^2}dV$ where $E$ is the solid region in $\mathbb{R}^3$ bounded by the surface $z=1$ and $z=x^2+y^2$.

$\begin{array}{cl} \iiint_E\sqrt{x^2+y^2}dV & =\int_0^1 {\int_0^{2\pi } {\int_0^{\sqrt z } {\sqrt {z^2 } r} dr} d\theta } dz \\ & =\int_0^1 {\int_0^{2\pi } {\int_0^z {\sqrt u } \cdot \frac{{du}}{2}} d\theta } dz \\ & =\frac{1}{2}\int_0^1 {\int_0^{2\pi } {\left. {\frac{{u^{\frac{3}{2}} }}{{\frac{3}{2}}}} \right|_{u = 0}^{u = z} } d\theta } dz\\&=\frac{1}{3}\int_0^1 {\int_0^{2\pi } {z^{\frac{3}{2}} } d\theta } dz\\&=\frac{1}{3}\int_0^1 {z^{\frac{3}{2}} \cdot 2\pi } dz = \frac{{2\pi }}{3} \cdot \left. {\frac{{z^{\frac{5}{2}} }}{{\frac{5}{2}}}} \right|_{z = 0}^{z = 1} = \frac{{4\pi }}{{15}} ∎ \end{array}$

Evaluate the line integral $\int_C {(ye^x + \sin y)dx} + (e^x + x\cos y)dy$ along the curve $C:\textbf{r}(t) = (t^2 + 1)\textbf{i} + (t^2 – 1)\textbf{j},\ 0\leq t\leq 2$. Hint: you may find a potential function $f$ of the vector field $\textbf{F}(x,y) = (ye^x + \sin y)\textbf{i} + (e^x + x\cos y)\textbf{j}$, i.e. find $f$ so that $\textbf{F}=\nabla f$.

$\because\ \textbf{F}=\nabla f=(f_x,f_y)$
$\therefore\ \left\{ \begin{array}{l} f_x = ye^x + \sin y \Rightarrow f = ye^x + x\sin y + \rho (y) \Rightarrow f_y = e^x + x\cos y + \rho'(y) \\ f_y = e^x + x\cos y \end{array} \right.$
$\Rightarrow \rho'(y)=0$
$\Rightarrow \rho (y) = C \Rightarrow f = ye^x + x\sin y + C$

$\begin{array}{cl} \int_C\textbf{F}d\textbf{r} & =f(5,3)-f(1,-1) \\ & =(3e^5 + 5\sin 3) – ( – 1 \cdot e^1 + 1 \cdot \sin ( – 1)) \\ & =5\sin 3 + \sin 1 + 3e^5 + e ∎\end{array}$