Question

lim1/(sinx)^2-(cosx)^2/x^2 x趋向于0

Answer 1

x趋于0时 (cosx)^2=1
所以 为 lim 1/(sinx)^2-1/x^2=0

lim[1/sinx^2-1/x^2]
=(x^2-sinx^2)/x^2*sinx^2
=(x^2-sinx^2)/x^4;
用洛必达法则求导
=(2x-sin2x)/4x^3
=(2-2cos(2x))/12x^2
=4sin(2x)/24x
=8x/24x
=1/3

Answer 2

=lim<x→0>[(x²-sin²xcos²x)/(sin²x·x²)]
=lim<x→0>[(x²- 1/4 sin²2x)/(x^4)]
=lim<x→0>[x²- 1/8 (1-cos4x)]/(x^4)
=lim<x→0>[2x- 1/8 (sin4x·4)]/(4x³)
=lim<x→0>[2x- 1/2 ·sin4x]/(4x³)
=lim<x→0>[2- 2 cos4x]/(12x²)
=lim<x→0>[2(1- cos4x)]/(12x²)
=lim<x→0>[2· 1/2 (4x)²]/(12x²)
=lim<x→0> 16/12
=4/3