微积分学习笔记二

1、罗尔中值定理:设函数y=f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=f(b),那么在(a,b)内至少存在一点ξ,使得

\[f^{^{'}}\left (\varepsilon \right )=0\]

证明:因为连续,所以在[a,b]上存在最大最小值,设为M,m。

(1)若M=m,则f(x)为定值,所以导数恒为0。

(2)否则,由于f(a)=f(b),所以M和m不可能都在端点处取得。设f(c)=M,c∈(a,b)。因为f(c)=M是最大值,所以对$\Delta x\neq 0$有

\[f\left(c+\Delta x \right )-f\left(c \right )\leq 0,c+\Delta x \in \left(a,b \right )\]

所以,当$\Delta x>0$时,

\[\frac{f\left(c+\Delta x \right )-f\left(c \right )}{\Delta x}\leq0\]

因为导数存在,所以

\[f^{^{'}}\left(c \right )=\lim_{\Delta x\rightarrow 0^{^{+}}}\frac{f\left(c+\Delta x \right )-f\left(c \right )}{\Delta x}\leq0\]

同理$\Delta x<0$时

\[f^{^{'}}\left(c \right )=\lim_{\Delta x\rightarrow 0^{^{-}}}\frac{f\left(c+\Delta x \right )-f\left(c \right )}{\Delta x}\geq 0\]

所以$f^{^{'}}\left(c \right )=0$。因此可取ξ=c。


2、拉格朗日中值定理:设函数f(x)在[a,b]上连续,在(a,b)内可导,则在(a,b)内至少存在一点ξ,满足

\[f^{^{'}}\left(\xi \right )=\frac{f\left(b \right )-f\left(a \right )}{b-a}\]

证明:定义函数$\varphi (x)$,

$\varphi \left(x \right )=f\left(x \right )-[f\left(a \right )+\frac{f(b)-f(a)}{b-a}(x-a) ]$


那么有$\varphi (a)$=$\varphi (b)$=0,由罗尔定理,(a,b)存在一点ξ使得$\varphi^{^{'}} \left(\xi\right )=0$,即


\[\varphi^{^{'}} \left(\xi \right )=f^{^{'}}\left(\xi \right )-\frac{f\left(b \right )-f\left(a \right )}{b-a}=0\]

所以\[f^{^{'}}\left(\xi \right )=\frac{f\left(b \right )-f\left(a \right )}{b-a}\]

3、柯西中值定理:设f(x),g(x)在[a,b]上连续,在(a,b)内可导,且$g^{^{'}}\left(x \right )\neq0 (a<x<b)$,则在(a,b)内存在一点ξ,满足

\[\frac{f\left(b \right )-f\left(a \right )}{g\left(b \right )-g\left(a \right )}=\frac{f^{^{'}}\left(\xi \right )}{g^{^{'}}\left(\xi \right )},\xi\in(a,b)\]

证明:由于$g^{^{'}}\left(x \right )\neq0(a<x<b )$,所以g(a)!=g(b)。否则与罗尔定理矛盾。定义函数$\varphi (x)$,

$\varphi (x)=f(x)-[f\left(a \right )+\frac{f\left(b \right )-f\left(a \right )}{g\left(b \right )-g\left(a \right )} [g(x)-g(a) ]]$


由于$\varphi (a)$=$\varphi (b)$=0,所以根据罗尔定理,在(a,b)内存在ξ,使得

\[\varphi^{^{'}} \left(\xi \right )=f^{^{'}}\left(\xi \right )-\frac{f\left(b \right )-f\left(a \right )}{g\left(b \right )-g\left(a \right )}g^{^{'}}\left(\xi \right )=0\]



\[\frac{f\left(b \right )-f\left(a \right )}{g\left(b \right )-g\left(a \right )}=\frac{f^{^{'}}\left(\xi \right )}{g^{^{'}}\left(\xi \right )},\xi\in(a,b)\]

4、洛必达法则1:设函数f(x),g(x)满足条件:

$(1)\lim_{x\rightarrow a }f\left(x \right )=0,\lim_{x\rightarrow a }g\left(x \right )=0$

(2)在点a的某邻域内(a可除外)可导,且$g^{^{'}}\left(x \right )\neq0$

$(3)\lim_{x\rightarrow a}\frac{f^{^{'}}\left(x \right )}{g^{^{'}}\left(x \right )}=A$(或者oo)

那么

$\lim_{x\rightarrow a}\frac{f\left(x \right )}{g\left(x \right )}=\lim_{x\rightarrow a}\frac{f^{^{'}}\left(x \right )}{g^{^{'}}\left(x \right )}=A$(或者oo)

证明:若f(x) 和g(x)在a处连续,那么f(a)=g(a)=0,否则我们可以人为定义f(a)=g(a)=0,使得f(x)和g(x)在a处连续。设x为a附近一点,那么在以x和a为端点的区间上,f(x)和g(x)满足柯西中值定理,所以在(x,a)之间存在ξ,使得

\[\frac{f\left(x \right )}{g\left(x \right )}=\frac{f\left(x \right )-f\left(a \right )}{g\left(x \right )-g\left(a \right )}=\frac{f^{^{'}}\left(\xi \right )}{g^{^{'}}\left(\xi \right )},\xi\in(x,a)\]

所以当x->a时,ξ->a,求极限得:

$\lim_{x\rightarrow a}\frac{f\left(x \right )}{g\left(x \right )}=\lim_{\xi \rightarrow a}\frac{f^{^{'}}\left(\xi \right )}{g^{^{'}}\left(\xi \right )}=\lim_{x \rightarrow a}\frac{f^{^{'}}\left(x \right )}{g^{^{'}}\left(x \right )}$

5、洛必达法则2:设函数f(x),g(x)满足条件:

$(1)\lim_{x\rightarrow a }f\left(x \right )=oo,\lim_{x\rightarrow a }g\left(x \right )=oo$

(2)在点a的某邻域内(a可除外)可导,且$g^{^{'}}\left(x \right )\neq0$

$(3)\lim_{x\rightarrow a}\frac{f^{^{'}}\left(x \right )}{g^{^{'}}\left(x \right )}=A$(或者oo)

那么

$\lim_{x\rightarrow a}\frac{f\left(x \right )}{g\left(x \right )}=\lim_{x\rightarrow a}\frac{f^{^{'}}\left(x \right )}{g^{^{'}}\left(x \right )}=A$(或者oo)


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