lim(x->0）[x ln(x)]这个极限怎么求?

lim(x->0）[x ln(x)]这个极限怎么求? lim(x趋于0）(ln(1+x)^1/x) lim(x趋近0)ln(x/sinx) lim(x→0)ln（x^2+1）等于 求极限:lim(x->0) ln(sinx/x)/(x*x) lim ln(e^sinx) x趋向0 lim→0+ lnx ln(1+X) lim(x趋于0+)(ln(xln a)ln(ln ax/ln(x/a))),其中a>1 lim(x->0)ln(1+2x)/e^x-1 求极限lim Ln(1+x) /x > .< x→0 x趋向0 lim [ ln (1-x) / (e ^ x-1 ) ] lim(ln(1+x)/x) x趋向0 lim(x->0)(ln(1+x)-x)/(cosx-1)=? lim(x趋向0)ln(1+sin x)/x^2 lim(x趋于0)[ln(1+2x)]/x lim(x->0)(ln(1+x))/x 不用导数 lim(x→0)tan(x)ln(1+x)=? f(x)=ln(x+1),lim(x->0)