函数连续性
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f(x)= [g(x)-e^(-x)]/x ; x≠0
= 0 ; x=0
(1)
lim(x->0) f(x)
= lim(x->0)[g(x)-e^(-x)]/x (0/0)
= lim(x->0)[g'(x)+e^(-x)]
=g'(0) +1
=0
=f(0)
x=0, f(x) 是连续
(2)
f'(0)
=lim(h->0) [ f(h) -f(0) ]/h
=lim(h->0) [g(h)-e^(-h)]/h^2 (0/0)
=lim(h->0) [g'(h)+e^(-h)]/(2h) (0/0)
=lim(h->0) [g''(h)-e^(-h)]/2
=(1/2) [g''(0) -1]
=> (-∞, ∞) f'(x) 是连续
= 0 ; x=0
(1)
lim(x->0) f(x)
= lim(x->0)[g(x)-e^(-x)]/x (0/0)
= lim(x->0)[g'(x)+e^(-x)]
=g'(0) +1
=0
=f(0)
x=0, f(x) 是连续
(2)
f'(0)
=lim(h->0) [ f(h) -f(0) ]/h
=lim(h->0) [g(h)-e^(-h)]/h^2 (0/0)
=lim(h->0) [g'(h)+e^(-h)]/(2h) (0/0)
=lim(h->0) [g''(h)-e^(-h)]/2
=(1/2) [g''(0) -1]
=> (-∞, ∞) f'(x) 是连续
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