Sichuan University 2010 years of Mathematical analysis of postgraduate examination questions

1. Calculate the following limits (7 points per sub-topic, total 28 points)

(1). $\displaystyle \lim\limits_{x\to 0} \frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin^{2}x}$.

(2) $\displaystyle \lim\limits_{n\to \infty} \left (\frac{1^{p}+2^{p}+\cdots +n^{p}}{n^{p}}-\frac{n}{p+1}\right) $,$ ( P\in N,p\ge 1) $.

(3) $\displaystyle \lim\limits_{x\to +\infty}\frac{\displaystyle x^{2\ln x}-x}{(\ln x) ^{x}+x}$.

(4) $\displaystyle \lim\limits_{n \to \infty}\sqrt[n]{1+a^{n}+\sin^{2}n},a>0$.

2. Calculate the integral (8 points for each small question, total 40 points)

(1). $\displaystyle \int\frac{\sin x \cos^{3}x}{1+\cos^{2}x}dx$;

(2). $\displaystyle \oint_{l}\frac{(x-y) dx+ (X+4Y) dy}{x^{2}+4y^{2}}$, wherein $l$ is the unit circle $x^{2}+y^{2}=1$, take counterclockwise direction;

(3). $\displaystyle \iint_{\sigma} (2x+z) dydz+zdxdy$, where $\sigma$ is the surface $z=x^{2}+y^{2} (0\le Z\le 1) $ fetch the upper side;

(4). $\displaystyle g (\alpha) =\int_{1}^{+\infty}\frac{\arctan \alpha x}{x^{2}\sqrt{x^{2}-1}}dx$;

(5). Set $\displaystyle f (x) =\int_{1}^{x}\frac{\sin t}{t}dt$, $\displaystyle \int_{0}^{1}xf (x) dx$.

3. (12 points) $f (x) $ in $[0,+\infty) $ continuous, and $\displaystyle \lim\limits_{x\to +\infty}\left (f (x) +\sin x\right) =0$. Proof: $f (x) $ in $ [0,+\infty] $ on consistent continuous.

4. (10 points) make $u=f (z) $, where $z=z (x, y) $ is the implicit function determined by the equation $\displaystyle Z=x+y\varphi (z) $, and $f (z) $ and $\varphi (z) $ are any order-micro functions. Proof:

$$\frac{\partial ^{n}u}{\partial y^{n}}=\frac{\partial ^{n-1}}{\partial x^{n-1}}\left[\left (\varphi (z) \right) ^{n}\ Frac{\partial u}{\partial x} \right]$$

5. (15 points) Proof: If the function $f (x) $ within $ (0,+\infty) $ can be micro, and $\displaystyle \lim\limits_{x\to +\infty}f ' (x) =0$. $\displaystyle \lim\ Limits_{x\to +\infty}\frac{f (x)}{x}=0$.

6. (10 points) set $f (x) $ within $[a,b]$ and $f (a) =0$. Proof: $ $M ^{2}\le (b-a) \int_{a}^{b}\left[f ' (x) \right]^{2}dx$$ where $M =\sup\limits _{a\le x\le b}\left\{\left|f (x) \right|\right\}$.

7. (20 points)

Set $\displaystyle F_{n} (x) =n^{\alpha}xe^{-nx},n\in n$. When argument $\alpha $ why value

(1). function column $\displaystyle \{f_{n} (x) \}$ converge on $[0,1]$;

(2). function column $\displaystyle \{f_{n} (x) \}$ uniformly converge on $[0,1]$;

(3). $\displaystyle \int_{0}^{1}\lim\limits_{n\to \infty}f_{n} (x) Dx=\lim\limits_{n\to\infty}\int_{0}^{1}f_{n} (x) dx$.

8. (15 points) Proof: $$\displaystyle \iiint\limits_{\omega}\frac{dxdydz}{r}=\frac{1}{2}\iint\limits_{\partial \Omega}\cos \left (\overrightarrow{r}, \overrightarrow{n}\right) ds$$

Where $\omega$ is a single connected area in the $r^{3}$, $\partial \omega$ for its smooth boundary surface, $\overrightarrow{n}$ for $\partial \omega$ in points $ (x, y, Z) $ of the unit outer vector, $r = \sqrt{(\xi-x) ^{2}+ (\eta-y) ^{2}+ (\zeta-z) ^{2}},\overrightarrow{r}= (x-\xi) \overrightarrow{i}+ (Y-\eta) \ overrightarrow{j}+ (Z-\zeta) \overrightarrow{k}$ is the vector of the connection space midpoint $ (\xi,\eta,\zeta) $ to $ (x, y, z) $.

Sichuan University 2010 years of Mathematical analysis of postgraduate examination questions