求大神解解. lim(ln(1+x)),x→-1
求极限lim Ln(1+x) /x > .< x→0
求极限:lim(x→+∞)[ln(x+1)-lnx]
求极限lim(x趋向无穷大)ln(1+x)/x
求极限lim (n→+∞)ln(1+x^2)/ln(1+x^4)
已知f(x)=ln(1+x) 求lim(x→0) f(x)/x
求该函数的极限 (x→1)lim [ x/(x-1) -1/ln x]
求lim(x→0+)[x分之1-x²分之ln(x+1)]的极限?
求极限:lim(x→0)ln(1+x²)/ (sec x- cos x)
用洛必达法则求lim(x→0)x²分之x-ln(1+x)
lim*[ln(1+3X)]/sin4X {X->0}求极限
求极限lim(ln(1+e^x)),x->+∞
求 lim ln(1+x+2x^2)+ln(1-x+x^2)/secx-cosx