2道求lim的题没学过taylor,最好是有stepslim(x→0)(sin^2x/2)/x^2lim(x→0)tan^3(2x)/(x^3)

Question

2道求lim的题
没学过taylor,最好是有steps
lim(x→0) (sin^2 x/2)/x^2
lim(x→0) tan^3(2x)/(x^3)

Answer 1

答：
这个用不着泰勒展开,用到重要极限limx->0 sinx/x=1
1.
原式
=limx->0 [sin(x/2)]^2/(x/2)^2*(x/2)^2/x^2
令x/2=t,有t->0,所以sint/t=(sint)^2/t^2=1
=limx->0 x^2/4x^2
=1/4
2.
原式
=limx->0 [sin(2x)]^3/(2x)^3*(2x)^3/[(x^3)[cos(2x)]^3]
与上题同理
=limx->0 8/cos(2x)^3
=8/1
=8