已知函数f(x)f(x)f(x)在点x=0x = 0x=0处连续,且当x≠0x\neq 0x=0时,函数f(x)=2−1x2f(x)=2^{-\dfrac{1}{x^{2}}}f(x)=2−x21,则函数值f(0)f(0)f(0)=( ).
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