利用等价无穷小和L'Hospital's Rule 即可
lim(x->0) (e^x-e^sinx ) / [ (tanx )^2 * ln(1+2x)]
=lim(x->0) e^x(e^(x-sinx)-1 ) / [ (tanx )^2 * ln(1+2x)]
=lim(x->0) (x-sinx ) / [ (x)^2 * 2x)]
=lim(x->0) (1-cosx)/(6x^2)
=lim(x->0) [(x^2)/2]/(6x^2)
=1/12
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