Question

求极限(x-->0),(根号(1-X^2))^(1/x) 要用洛必达定理.

Answer 1

【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