设$f(x)$具有一阶连续导数, 且
[
limlimits_{x o0}left(frac{{
m e}^x-1}{x^2}+frac{f(x)}{x}
ight)=3,
]
求$f(0)$以及$f'(0)$.
解: 由
[
limlimits_{x o0}left(frac{{
m e}^x-1}{x^2}+frac{f(x)}{x}
ight)
=limlimits_{x o0}frac{frac{{
m e}^x-1}{x}+f(x)}{x}
=3
]
可知
[
limlimits_{x o0}left(frac{{
m e}^x-1}{x}+f(x)
ight)=0,
]
即有[f(0)=limlimits_{x o0}f(x)=-limlimits_{x o0}frac{{
m e}^x-1}{x}=-1.]
注意到
[
lim_{x o0}frac{{
m e}^x+f(x)+xf'(x)}{2x}
=lim_{x o0}left(frac{{
m e}^x-1}{2x}+frac{f(x)-f(0)}{2x}+frac{1}{2}f'(x)
ight)
=frac{1}{2}+f'(0),
]
可知
[
lim_{x o0}frac{{
m e}^x-1+xf(x)}{x^2}=frac{1}{2}+f'(0)=3,
]
因此有(f'(0)=frac{5}{2}.)
也可以考虑泰勒展开, 由于$x o0$时,
[
{
m e}^x=1+x+frac{1}{2}x^2+o(x^2),
]
[
f(x)=f(0)+f'(0)x+o(x),
]
可得
egin{align*}
&lim_{x o0}frac{{
m e}^x-1+xf(x)}{x^2}\
=&lim_{x o0}frac{x+frac{1}{2}x^2+o(x^2)-x+f'(0)x^2+o(x^2)}{x^2}\
=&frac{1}{2}+f'(0)=3,
end{align*}
因此有(f'(0)=frac{5}{2}.)