Question

【高数】利用两个重要极限求函数极限

Answer 1

解：lim(x->0)[(tanx-sinx)/x³]=lim(x->0)[(sinx/cosx-sinx)/x³]
=lim(x->0)[(1/cosx)(sinx/x)((1-cosx)/x²)]
=lim(x->0)[((1/2)/cosx)(sinx/x)(sin(x/2)/(x/2))²]
(应用余弦倍角公式)
=lim(x->0)[(1/2)/cosx]*lim(x->0)[(sinx/x)]*[lim(x->0)(sin(x/2)/(x/2))]²
=(1/2)*1*1²
(应用重要极限lim(z->0)(sinz/z)=1)
=1/2；
lim(x->1)[(1-x)tan(πx/2)=lim(y->0)[ytan(π/2-πy/2)]
(令y=1-x)
=lim(y->0)[ycot(πy/2)]
(应用诱导公式)
=lim(y->0)[(y/sin(πy/2))cos(πy/2)]
=lim(y->0)[((πy/2)/sin(πy/2))(2cos(πy/2)/π)]
=lim(y->0)[(πy/2)/sin(πy/2)]*lim(y->0)[2cos(πy/2)/π]
=1*(2/π)
(应用重要极限lim(z->0)(sinz/z)=1)
=2/π；
lim(x->0)[(1-2x)^(1/x)]=lim(x->0)[(1+(-2x))^((1/(-2x))(-2))]
=lim(x->0)[((1+(-2x))^((1/(-2x)))^(-2)]
=[lim(x->0)((1+(-2x))^(1/(-2x)))]^(-2)
=e^(-2)
(应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=1/e²；
lim(n->∞)[(1+2/5^n)^(5^n)]=lim(n->∞)[(1+2/5^n)^((5^n/2)*2)]
=[lim(n->∞)((1+2/5^n)^(5^n/2)]²
=e²
(应用重要极限lim(z->0)[(1+z)^(1/z)]=e)。

Answer 2

“凑”重要极限的形式而已
1、分子化成tanx(1-cosx)，整个式子化为tanx/x×(1-cosx)/x^2
2、tan(πx/2)写成1/tan[π/2×(1-x)]，把1-x添加一个π/2即可
3、1/x写成1/(-2x)×(-2)
4、5^x＝5^x/2×2