设 函数 f(x)在x=2处可导,且f(2)的导数=1求: lim f(2+h)—f(2—h)/2h h→0
问题描述:
设 函数 f(x)在x=2处可导,且f(2)的导数=1求: lim f(2+h)—f(2—h)/2h h→0
答
lim {[f(2+h)-f(2-h)]/(2h)}
=lim {[f(2+h)-f(2)]/(2h)+f(2)/(2h)-[f(2-h)-f(2)]/(2h)-f(2)/(2h) }
=lim {[f(2+h)-f(2)]/(2h)-[f(2-h)-f(2)]/(2h)}
=(1/2)lim {[f(2+h)-f(2)]/h+[f(2)-f(2-h)]/h}
=(1/2)lim [f'(2)+[f'(2)]
=lim f'(2)
=1
答
lim f(2+h)—f(2—h)/2h =lim ((f(2+h)—f(h)+ f ( h )—f(2—h))/2h ))
=lim((f(2+h)—f(h)/(2h)) +lim(f ( h )—f(2—h))/(2h ))
=(1/2)*lim((f(2+h)—f(h)/h) +lim(f ( h )—f(2—h))/h )
=(1/2)*(f(2)的导数 +f(2)的导数 )
=(1/2)*2
=1
答
lim f(2+h)—f(2—h)/2h =lim[ f(2+h)-f(2)/2h—(f(2—h)-f(2))/2h ]
=f'(2)/2+f'(2)/2=1