lim(x→0)(x/sinx)=1 lim(x→0)ln(1+2x)^(1/2x)=1/2 lim(x→0)(1+2x)/x=1/2 lim(x→0)2x/x=1所以原式等于6(x→0)(x/sinx)lim(x→0)(x/sinx)=6*1*1=6
lim(x→0)[6(x/sinx)ln(1+2x)^(1/2x)]谢谢了,
1个回答
相关问题
-
x→0时 lim[1/ln(1+x^2)-1/(sinx)^2] =lim[(sinx)^2-ln(1+x^2)]/(s
-
求极限lim(x→0)[2sinx+x^2(sin1/x)]/ln(1+x)
-
lim((sinx-ln(1+x^2)^1/2)/x^4) x->0求解 呵呵~
-
lim(cos(1/x)+2/sinx-1/ln(1+x)) x趋近于0
-
求极限,lim(x->0) (e^x-e^sinx ) / [ (tanx )^2 * ln(1+2x)]
-
求极限,lim(x->0) (1-2sinx)^(3/x)lim(n->+∞) (n!-4^n) / (6+ln(n)+
-
极限lim (1+x)ln(1+x)分之sinx = ? x→0
-
lim ln(sinx/x)=ln (lim(sinx/x))
-
lim [ln(1+ax^2)]/(x sinx) (x趋近于0负)
-
lim(x→0)[ln(1+x)]/x=lim(x→0)[ln(1+x)]^(1/x)