求极限.sin(1-x)/lnx(x趁于1）；(x^2+2x)/sinx(x趁于0）.用等价无穷小代换法求.

题目

求极限.sin(1-x)/lnx(x趁于1）；(x^2+2x)/sinx(x趁于0）.用等价无穷小代换法求.

答案

x->1 x-1-->0 sin(1-x)~1-x
lnx=ln(1+(x-1))~(x-1)
limsin(1-x)/lnx=-1
x->0 sinx~x
lim(x²+2x)/sinx=lim(x²+2x)/x=2