0),(根号(1-X^2))^(1/x) 要用洛必达定理.

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【罗必塔法则】
lim(x->0) [√(1-x^2)]^(1/x)
=lim(x->0) e^{ (1/x)ln[√(1-x^2)] }
=lim(x->0) e^{ 1/2 ln(1-x^2) /x }
= e^ { 1/2*lim(x->0)[ln(1-x^2)/x] }
罗必塔法则= e^{ 1/2*lim(x->0) [(-2x)/(1-x^2) /1 ]}
= e^0 = 1【重要极限其实更简洁】
lim(x->0) [(1-x^2)^(1/2)]^(1/x)
=lim(x->0) [1+(-x^2)]^(1/2x)
=lim(x->0) {[1+(-x^2)]^(-1/x^2)}^(-x/2)
= e^0= 1