作业帮【高数】利用两个重要极限求函数极限
来源：学生作业帮 编辑：拍题作业网作业帮 分类：数学作业 时间：2024/04/27 17:42:32
【高数】利用两个重要极限求函数极限
求一下几个函数的极限,结果我知道,怎么变形的.
求一下几个函数的极限,结果我知道,怎么变形的.
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).
=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).