Question

求极限lim(x→0){(2x-sin2x)/xsin^2x}

Answer 1

lim(x→0){(2x-sin2x)/xsin^2x}=lim(x→0){(2-2cos2x)/(sin^2x+2xsinxcosx)} （分子分母分别求导）
=lim(x→0){(4sin2x)/(2sinxcosx+sin2x+2xcos2x)}=lim(x→0){(8cos2x)/(4cos2x-4xsin2x+2cos2x)}
=8/6=4/3

Answer 2

x→0时,2x/sin2x=1
即x→0时,2x-sin2x=0
所以lim(x→0){(2x-sin2x)/xsin^2x} =0

Answer 3

lim(x→0){(2x-sin2x)/(x*sin^2x)}
= lim(x→0){(2x-sin2x)/(x*x^2*(sin^2x/x^2))}
= lim(x→0){(2x-sin2x)/(x*x^2) * lim(x→0){(sin^2x/x^2)}
= lim(x→0){(2-2cos2x)/(3x^2)
= lim(x→0){(4sin2x)/(6x)
= lim(x→0){(8cos2x)/(6)
= 4/3