求极限,lim(x->0) (e^x-e^sinx ) / [ (tanx )^2 * ln(1+2x)]

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)...