lim(1/ln(x+1)-1/x) x趋于零 用洛必达法则

lim[ln(1+x)+ln(1-x)]/(tanx)^2 为什么 lim ln[(1+1/x)^x]=ln e lim ln(1+x)/x 怎么推到lim ln(1+x)1/x lim((ln(1+x))/x)^(1/x) x->无穷大 lim(x->1)lnx*ln(x 1)等于多少是lim(x->1)ln(x)ln(x-1) lim→0+ lnx ln(1+X) lim(x趋于0+)(ln(xln a)ln(ln ax/ln(x/a))),其中a>1 计算：lim[1/x+1/(x^2)ln(1-x)] lim(1/ln(1+x)-1/x) 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(x->0)(ln(1+x)-x)/(cosx-1)=? lim ln(1+x)^ 1/x x趋近于0 lim(x趋于0）(ln(1+x)^1/x) 求极限lim Ln(1+x) /x > .< x→0 求极限lim(x趋向无穷大)ln(1+x)/x lim(ln(1+x)/x) x趋向0 lim(x趋向0)ln(1+sin x)/x^2