lim(x→0)(ln(1+x^2)/(sec-cosx))这题怎么解哈 大哥大姐帮帮忙啊

lim(x→0)(cosx)^(1/ln(1+x^2)) lim(x→0)ln（x^2+1）等于 lim→0+ lnx ln(1+X) lim(x→0) (ln cosx)/[ln(1+x^2)] 等于多少? 求极限lim Ln(1+x) /x > .< x→0 lim(x→0)tan(x)ln(1+x)=? lim(x→0)[ln(2+x)-ln2]/x lim(x->0)ln(1+2x)/e^x-1 lim(x趋向0)ln(1+sin x)/x^2 lim(x趋于0)[ln(1+2x)]/x 对数函数的极限 lim(x→0) [ln(1+x)-ln(1-x)]/x lim(x趋近于0)[1/ln(x+根号1+x^2)-1/ln(1+x)] lim(1/x)ln(1+x+x^2)(x→0)解答过程 求极限lim(x→0)[2sinx+x^2(sin1/x)]/ln(1+x) lim x→0[ln(cosx)]/x^2 怎么计算lim(x->0+)x^(1/2)*ln(sinx)? lim x趋于0 ln (1+2x)/sin3x lim[ln(1+x)+ln(1-x)]/(tanx)^2