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

Question

3.lim(x→0)tanx-sinx/x^3
4.lim(x→π)sin3x/x-π
5.lim(x→无穷大量）（1-1/2x)^x+2
知道答案，求过程！谢谢

Answer 1

解：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)。

追问

3. =3*lim(x→π)[sin(3(x-π))/(3(x-π))] (应用诱导公式) 这一部我不懂

追答

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)))

Answer 2

分子分母同时求导，直到代入x的值分母不为零

Answer 3

洛比达法则
分子分母同时求导：
直到可以直接带入为止