求下列极限：1.lim(x→2)(x+2/x-2) 2.lim(x→0)4x^3-2x^2+x^2-2x)

3。lim(x→0)[(tanx-sinx)/x^3]=lim(x→0)[sinx(1-cosx)/(x³cosx)]=lim(x→0){(sinx/x)*[sin(x/2)/(x/2)]²*[1/(2cosx)]}=[lim(x→0)(sinx/x)]*{lim(x→0)[sin(x/2)/(x/2)]}²*{lim(x→0)[1/(2cosx)]}=1*1²*(1/2) (应用重要极限lim(z->0)(sinz/z)=1)=1/2；4。lim(x→π)[sin(3x)/(x-π)]=lim(x→π)[sin(3(π+x-π))/(x-π)] =3*lim(x→π)[sin(3(x-π))/(3(x-π))] (应用诱导公式)=3*1 (应用重要极限lim(z->0)(sinz/z)=1)；5。lim(x→∞)[(1-1/(2x))^x+2]=lim(x→∞){[(1+1/(-2x))^(-2x)]^(-(x+2)/(2x))}=e^{lim(x→∞)[-(x+2)/(2x)]} (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)=e^{lim(x→∞)[-(1+2/x)/(2)]}=e^[-(1+0)/2] =e^(-1/2)。 再问： 3。 =3*lim(x→π)[sin(3(x-π))/(3(x-π))] (应用诱导公式) 这一部我不懂 再答： lim(x→π)[sin(3(π+x-π))/(x-π)]=lim(x→π)[3*sin(3(π+x-π))/(3(x-π))] (分子分母同乘3，且x=π+x-π) =3*lim(x→π)[sin(3π+3(x-π))/(3(x-π))] =3*lim(x→π)[sin(3(x-π))/(3(x-π))] (应用诱导公式,即sin(3π+3(x-π))=sin(3(x-π)))