Question

一道极限求值提.不用罗比达法则,求：当x无限趋近于0时, （x*sinx）/(1-cosx)的值额。。。是 xsinx不

Answer 1

0/0用洛必达法则,原式=(sinx+x*cosx)/sinx=(2cosx-x*sinx)/cosx=2 x无限趋近于0

修改:

x不等于0时,(x*sinx）/(1-cosx)

=(x*2sin(x/2)cos(x/2))/(2(sin(x/2))^2)

=(x*cos(x/2))/(sin(x/2))

故所求为limx->0 (x*cos(x/2))/(sin(x/2))

=2(limx->0 cos(x/2))/limx/2->0 (sin(x/2))/(x/2)=2

运用到重要极限:limx->0 sinx/x=1