lim(x->0+)(sinx/x)^(1/x)=?,我知道答案,
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lim(x->0+)(sinx/x)^(1/x)=?,我知道答案,
lim(x->0+)(sinx/x)^(1/x)=?,我知道答案,
lim(x->0+)(sinx/x)^(1/x)=?,我知道答案,
这是1^∞型,这是通用方法化为((1+0)^1/0)^(0*∞)=e^(0*∞)
所以这题=[(1+(sinx-x)/x)^(x/(sinx-x))]^((sinx-x)/x/x)=e^((sinx-x)/x^2)
然后对指数洛比达法则,(cosx-1)/2x,再洛比达得-sinx/2
所以最后答案e^0=1