当x趋于0时lim(1/x2-1/tanx2) 的极限,咋做
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利用等价无穷小,洛必达法则求解.
(x->0)lim(1/x^2-1/arctan^2(x))
=(x->0)lim(1/x^2-cos^2(x)/sin^2(x))
=(x->0)lim[sin^2(x)-x^2cos^2(x)]/[x^2sin^2(x)]
=(x->0)lim(sinx+xcosx)(sinx-xcosx)/x^4
=(x->0)lim[(sinx+xcosx)/x][(sinx-xcosx)/x^3]
=(x->0)lim[(sinx)/x+cosx][(sinx-xcosx)]'/[x^3]'
=(x->0)lim(1+1)[(cosx-cosx+xsinx)/(3x^2)]
=(x->0)lim(2/3)(sinx)/x
=2/3
(x->0)lim(1/x^2-1/arctan^2(x))
=(x->0)lim(1/x^2-cos^2(x)/sin^2(x))
=(x->0)lim[sin^2(x)-x^2cos^2(x)]/[x^2sin^2(x)]
=(x->0)lim(sinx+xcosx)(sinx-xcosx)/x^4
=(x->0)lim[(sinx+xcosx)/x][(sinx-xcosx)/x^3]
=(x->0)lim[(sinx)/x+cosx][(sinx-xcosx)]'/[x^3]'
=(x->0)lim(1+1)[(cosx-cosx+xsinx)/(3x^2)]
=(x->0)lim(2/3)(sinx)/x
=2/3
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