利用等价无穷小求极限：lim(x→0)(cosx+2sinx)^(1/x)
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lim(x→0) (cosx+2sinx)^(1/x)
=lim(x→0) [1+(cosx-1+2sinx)]^(1/简梁做x)
=lim(x→0) {[ 1 + (cosx-1+2sinx) ]^[1/(cosx-1+2sinx)]}^[(cosx-1+2sinx)(1/渣乎x)]
∵lim(x→0) {[ 1 + (cosx-1+2sinx) ]^[1/(cosx-1+2sinx)]} = e
∵lim(x→0) [(cosx-1+2sinx)(1/x)]
=lim(x→拦衡0) [cosx-1]/x + lim(x→0) 2sinx/x
=lim(x→0) [-x^2/2]/x + lim(x→0) 2x/x
= 0+2 = 2
= e^2
=lim(x→0) [1+(cosx-1+2sinx)]^(1/简梁做x)
=lim(x→0) {[ 1 + (cosx-1+2sinx) ]^[1/(cosx-1+2sinx)]}^[(cosx-1+2sinx)(1/渣乎x)]
∵lim(x→0) {[ 1 + (cosx-1+2sinx) ]^[1/(cosx-1+2sinx)]} = e
∵lim(x→0) [(cosx-1+2sinx)(1/x)]
=lim(x→拦衡0) [cosx-1]/x + lim(x→0) 2sinx/x
=lim(x→0) [-x^2/2]/x + lim(x→0) 2x/x
= 0+2 = 2
= e^2
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