求极限 要详细步骤谢谢

Question

如果能用文字详细说明追加50

Answer 1

解法一：（罗必达法）（1）原式=e^{lim(x->0)[ln(1-x)/x]} =e^{lim(x->0)[-1/(1-x)]} (0/0型极限，应用罗比达法则) =e^(-1) =1/e；（2）原式=e^{lim(x->0)[ln(1+2x)/x]} =e^{lim(x->0)[2/(1+2x)]} (0/0型极限，应用罗比达法则) =e^2 =e2；（3）原式=e^{lim(n->∞)[2nln((n+2)/(n+1))]} =e^{lim(n->∞)[2ln((1+2/n)/(1+1/n))/(1/n)]} =e^{lim(x->0)[2ln((1+2x)/(1+x))/x]} (令x=1/n) =e^{lim(x->0)[2/((1+2x)(1+x))]} (0/0型极限，应用罗比达法则) =e^2 =e2。解法二：（重要极限法）（1）原式=lim(x->0){[(1+(-x))^(1/(-x))]^(-1)} ={lim(x->0)[(1+(-x))^(1/(-x))]}^(-1) =e^(-1) (应用重要极限lim(z->0)[(1+z)^(1/z)]=e) =1/e；（2）原式=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) =e2；（3）原式=lim(n->∞){[(1+1/(n+1))^(1/(n+1))]^(2n/(n+1))} ={lim(n->∞)[(1+1/(n+1))^(1/(n+1))]}^[lim(n->∞)(2n/(n+1))] =e^[lim(n->∞)(2/(1+1/n))] (应用重要极限lim(z->0)[(1+z)^(1/z)]=e) =e^2 =e2。