dailyMath

dailyMath

Today’s Question

求下列极限:

  1. \[ \lim_{x \rightarrow 0} \frac{(1+\cos x)^x-2^x}{\sin^3 x}\]
  2. \[ \lim_{x \rightarrow 1} \frac{x^x-x}{\ln x -x +1} \]

Yesterday’s Question

设\(f(x)\)在\( [0,1] \)二阶连续可微,且满足 \( f”(x) \le 0 \) , \( 0 \le f(x) \le 1 \) , \(f(0)=f(1) = 0\)

求证:平面曲线 \( y=f(x) , x \in [0,1] \) 的弧长 \( s \le 3\)

The solution of Yesterday’s Question

由于\( f(x) \)在\( [0,1] \)连续,从而存在\( x_0 \in [0,1] \)使\(f(x)\)在\(x=x_0\)处取最大值。
(1)如果 \( f(x_0)=0 \),则\(f(x)=0, x\in[0,1]\),所以这段曲线的弧长\(s=1 \le 3 \)得证
(2)如果\( f(x_0)>0 \),有\( f'(x_0) = 0 \), 又因为\( f”(x) \le 0 \),所以有\( f'(x) \ge 0, x \in [0,c] \), \( f'(x) \le 0, x \in [c,1] \)

从而:\[ \begin{align}
s & = \int_0^1 {\sqrt{1+|f'(x)|^2}dx}\\
~ &= \int_0^c {\sqrt{1+|f'(x)|^2}dx} +\int_c^1 {\sqrt{1+|f'(x)|^2}dx} \\
~ & \le \int_0^c { (1+f'(x) )dx} +\int_c^1 {(1-f'(x))dx} \\
~ & = [x+f(x)]^c_0 +[x-f(x)]^1_c \\
~&= 1+2f(c) \le 3\\
\end{align}\]

故得证。

证明:函数\(f(x,y) = \frac{xy}{x^2+y^2}\)在\((0,0)\)点的偏导数存在,且在\((0,0)\)处不可微

求不定积分:\[ \int{(\frac{\arctan x}{x-\arctan x})^2 dx}\]

提示: \[ -\frac{v’}{v^2}dv = d(\frac{1}{v})\]

\[ \begin{align} ~& \int{(\frac{\arctan x}{x-\arctan x})^2 dx} \\
&= \int{\frac{t^2}{(\tan t – t)^2} d(tan(t))} \\
&= \int{\frac{t^2}{(\cos t( \tan t – t))^2} d(t)} \\
&= \int{\frac{t^2}{( \sin t – t \cos t)^2} d(t)} \\
&= \int{\frac{t^2}{( \sin t – t \cos t)^2} d(t)} \\
&= \int{\frac{t}{\sin t}\frac{ t \sin t}{( \sin t – t \cos t)^2} d(t)} \\
&= \int{\frac{t}{\sin t}d(\frac{1}{\sin t – t \cos t})} \\
&= \frac{t}{\sin t} \frac{1}{\sin t – t \cos t} – \int{\frac{1}{\sin t – t \cos t} d(\frac{t}{\sin t})}\\
&= \frac{t}{\sin t} \frac{1}{\sin t – t \cos t} – \frac{\cos t}{\sin t} + C\\
&= \frac{x \arctan x}{x – \arctan x} + C\\
\end{align}
\]