(1) f(x)=e−1x2f_{(x)} = e^{-\frac{1}{x^2}}f(x)=e−x21 f(0)′=limh→0f(h)−f(0)h−0=limh→0e−1h2hf'_{(0)} = \lim_{h\to0}\frac{f_{(h)}-f_{(0)}}{h-0} = \lim_{h\to0}\frac{e^{-\frac{1}{h^2}}}{h}f(0)′=limh→0h−0f(h)−f(0)=limh→0he−h21 指数関数は1次関数より収束が速いので f(0)′=0よって0f'_{(0)} = 0 \qquad \text{よって} \qquad 0f(0)′=0よって0 (2) f(0)′′=limh→0f(h)′−f(0)′h−0f''_{(0)} = \lim_{h\to0}\frac{f'_{(h)}-f'_{(0)}}{h-0}f(0)′′=limh→0h−0f(h)′−f(0)′ (1) と同様に考えて f(0)′′=0f''_{(0)} = 0f(0)′′=0