求极限 ~~~~~~~~~~~~~~
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x->0
sinx ~ x-(1/6)x^3
sin2x ~ 2x - (4/3)x^3
x- (1/2) sin2x ~ (2/3)x^3
---------------------------
lim(x->0) [1/(xsinx) - cosx/x^2 ]
=lim(x->0) (x-sinx.cosx) /(x^2.sinx)
=lim(x->0) [ x-(1/2)sin(2x) ] /(x^2.sinx)
=lim(x->0) (2/3)x^3 /(x^3)
=2/3
sinx ~ x-(1/6)x^3
sin2x ~ 2x - (4/3)x^3
x- (1/2) sin2x ~ (2/3)x^3
---------------------------
lim(x->0) [1/(xsinx) - cosx/x^2 ]
=lim(x->0) (x-sinx.cosx) /(x^2.sinx)
=lim(x->0) [ x-(1/2)sin(2x) ] /(x^2.sinx)
=lim(x->0) (2/3)x^3 /(x^3)
=2/3
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