特殊函数求极限和求导
特殊函数求极限和求导:
1、不用洛比他法则,利用假设极限值,转化函数表达式,利用对数求极限: a x -1求极限lim x →0x
a x -1=b ,则有lim (1+bx ) =lim a x ,取对数,lim ln(1+bx ) =lim x ln a ,解:设lim x →0x →0x →0x →0x →0x
lim b ln(1+bx ) x →01bx a x -1=b =ln a ,即lim =ln a x →0x
ln(1+
例如已知,lim x →0f (x ) ) =3,求lim f (x ) 。 x →0x 22x -1
f (x ) x 2
3(2x -1) =lim (e -1) , 解:由题意lim 2⨯x →0x sin x x →0
f (x ) e 3(2-1) -1sin x , lim 2=lim ⨯lim =m (假设)x →0x x →0x →0x x
则有:lim mx ln(1+mx ) x →01mx x =3lim (2x -1) , x →0
lim m ln(1+mx ) x →01mx f (x ) 2x -1=m =3lim =3ln 2。即lim 2=3ln 2 x →0x x →0x
a x -1=ln a 。 这里就利用了lim x →0x
2 、对特殊函数的求导:
d ((f (x )) g (x ) ) ,设y =(f (x )) g (x ) ,取对数,ln y =g (x ) ln(f (x )) ,两边分别求导, dx
1g (x ) g (x ) y '=g '(x ) ln(f (x )) +⋅f '(x ) ,y '=(g '(x ) ln(f (x )) +⋅f '(x )) ⋅y y f (x ) f (x )
(1+x ) -e , x →0x
11x 例如:利用洛必达法则求lim 1
x g (x ) x -(1+x ) ln(1+x ) 令y =(1+x ) ,y '=(g '(x ) ln(f (x )) +⋅f '(x )) ⋅y =⋅(1+x ) x 2f (x ) x (1+x )
1
x (1+x ) -e 分子分母分别求导x -(1+x ) ln(1+x ) 分子分母分别求导lim =lim y '=e lim =2x →0x →0x x x (1+x )
1--ln(1+x ) 分子分母分别求导=-e e lim =e lim x →02x +3x 2x →02+6x 2