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
ight )=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$Delta x
eq 0$有
[fleft(c+Delta x
ight )-fleft(c
ight )leq 0,c+Delta x in left(a,b
ight )]
所以，当$Delta x>0$时，
[frac{fleft(c+Delta x
ight )-fleft(c
ight )}{Delta x}leq0]
因为导数存在，所以
[f^{^{'}}left(c
ight )=lim_{Delta x
ightarrow 0^{^{+}}}frac{fleft(c+Delta x
ight )-fleft(c
ight )}{Delta x}leq0]
同理$Delta x<0$时
[f^{^{'}}left(c
ight )=lim_{Delta x
ightarrow 0^{^{-}}}frac{fleft(c+Delta x
ight )-fleft(c
ight )}{Delta x}geq 0]
所以$f^{^{'}}left(c
ight )=0$。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
ight )=frac{fleft(b
ight )-fleft(a
ight )}{b-a}]
证明：定义函数$varphi (x)$，
$varphi left(x
ight )=fleft(x
ight )-[fleft(a
ight )+frac{f(b)-f(a)}{b-a}(x-a) ]$
那么有$varphi (a)$=$varphi (b)$=0,由罗尔定理，(a,b)存在一点ξ使得$varphi^{^{'}} left(xi
ight )=0$，即
[varphi^{^{'}} left(xi
ight )=f^{^{'}}left(xi
ight )-frac{fleft(b
ight )-fleft(a
ight )}{b-a}=0]
所以[f^{^{'}}left(xi
ight )=frac{fleft(b
ight )-fleft(a
ight )}{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$g^{^{'}}left(x
ight )
eq0 (a<x<b)$，则在(a,b)内存在一点ξ,满足
[frac{fleft(b
ight )-fleft(a
ight )}{gleft(b
ight )-gleft(a
ight )}=frac{f^{^{'}}left(xi
ight )}{g^{^{'}}left(xi
ight )},xiin(a,b)]
证明：由于$g^{^{'}}left(x
ight )
eq0(a<x<b )$，所以g(a)!=g(b)。否则与罗尔定理矛盾。定义函数$varphi (x)$,
$varphi (x)=f(x)-[fleft(a
ight )+frac{fleft(b
ight )-fleft(a
ight )}{gleft(b
ight )-gleft(a
ight )} [g(x)-g(a) ]]$
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
ight )=f^{^{'}}left(xi
ight )-frac{fleft(b
ight )-fleft(a
ight )}{gleft(b
ight )-gleft(a
ight )}g^{^{'}}left(xi
ight )=0]
即
[frac{fleft(b
ight )-fleft(a
ight )}{gleft(b
ight )-gleft(a
ight )}=frac{f^{^{'}}left(xi
ight )}{g^{^{'}}left(xi
ight )},xiin(a,b)]
4、洛必达法则1：设函数f(x),g(x)满足条件:
$(1)lim_{x
ightarrow a }fleft(x
ight )=0,lim_{x
ightarrow a }gleft(x
ight )=0$
(2)在点a的某邻域内（a可除外）可导，且$g^{^{'}}left(x
ight )
eq0$
$(3)lim_{x
ightarrow a}frac{f^{^{'}}left(x
ight )}{g^{^{'}}left(x
ight )}=A$(或者oo)
那么
$lim_{x
ightarrow a}frac{fleft(x
ight )}{gleft(x
ight )}=lim_{x
ightarrow a}frac{f^{^{'}}left(x
ight )}{g^{^{'}}left(x
ight )}=A$(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{fleft(x
ight )}{gleft(x
ight )}=frac{fleft(x
ight )-fleft(a
ight )}{gleft(x
ight )-gleft(a
ight )}=frac{f^{^{'}}left(xi
ight )}{g^{^{'}}left(xi
ight )},xiin(x,a)]
所以当x->a时，ξ->a，求极限得：
$lim_{x
ightarrow a}frac{fleft(x
ight )}{gleft(x
ight )}=lim_{xi
ightarrow a}frac{f^{^{'}}left(xi
ight )}{g^{^{'}}left(xi
ight )}=lim_{x
ightarrow a}frac{f^{^{'}}left(x
ight )}{g^{^{'}}left(x
ight )}$
5、洛必达法则2：设函数f(x),g(x)满足条件:
$(1)lim_{x
ightarrow a }fleft(x
ight )=oo,lim_{x
ightarrow a }gleft(x
ight )=oo$
(2)在点a的某邻域内（a可除外）可导，且$g^{^{'}}left(x
ight )
eq0$
$(3)lim_{x
ightarrow a}frac{f^{^{'}}left(x
ight )}{g^{^{'}}left(x
ight )}=A$(或者oo)
那么
$lim_{x
ightarrow a}frac{fleft(x
ight )}{gleft(x
ight )}=lim_{x
ightarrow a}frac{f^{^{'}}left(x
ight )}{g^{^{'}}left(x
ight )}=A$(或者oo)