lim(x->0)(1-e^1/x)/(1+e^1/x)不存在,此题如何解释左右极限不等?证明lim(x->0)(1-e^1/x)/(1+e^1/x)不存在证：原式=lim(x->0){[2-1-e^(1/x)]/[1+e^(1/x)]}=lim(x->0){2/[1+e^(1/x)]-1}∵右极限=lim(x->0+){2/[1+e^(1/x)]-1}=-1左极限=lim

lim(x->0+) e^(1/x) x趋近于0 lim(x+e^x)^1/x lim(x→0)e^x-x-1/x^2 lim (e-(1+x)^(1/x))/x lim(x+e^3x)^1/x lim(lnx)-1/(x-e) lim趋向0时,e^x-e^-1/X 求极限lim (e^1/x+e)tanx/x(e^1/x-e) x趋于0^+ lim(x->0)ln(1+2x)/e^x-1 x趋向0 lim [ ln (1-x) / (e ^ x-1 ) ] lim(1+e^1/x)^x,x趋向于+0 lim趋近0 [(1+x)^1/x-e]/x lim[(1-e^(-x))^1/2]/x x趋于0 计算极限lim{x~0}(e^x-1)/x 求lim x趋向于0(e^x-x-1) lim(x→0) (e^(-1/x^2))/x^100 lim(x→0)(e^x-cosx)/x=1? lim(x→0)(2-e^x)^(1/x)