Question

x趋向于零时 [(1+sinx)^x - 1]/x 求极限

Answer 1

(1+sinx)^x - 1

= e^ [ ln (1+sinx)^x ] - 1

= e^ [ x*ln(1+sinx) ] - 1

x*ln(1+sinx)

【等价无穷小 e^x -1 x 】

所以原式 = lim(x→0) x*ln(1+sinx) / x

= lim(x→0) ln(1+sinx)

= 0