Question

高数上 求极限 很急很关键

距离交作业还有2小时
指点一下思路也可以啊！ 尤其是后两道

Answer 1

1. lim (x-> 0) e^(sin^-2 x) / (x^(-2)),用洛必达法则， = lim (x->0) e^(sin^-2 x) * (-2sin^-3 x)*cosx / (-2x^(-3)) = 无穷大

2. =e^(lim(x->0)sinx* lnx) = e^0 = 1

3. = lim(x->1) x(1-x)/ (1-x + lnx) = lim(x->1)x(1-x)/(1-x) = 1

4. =e^( lim(n->无穷大)nx*ln(.../n)) = e^(lim(n->无穷大)x*n*ln(1/n)) = e^x