大一高等数学求极限1.[㏑(x-π/2)]/tan x 当x趋于π/2时的极限2 cotx -1/x当x趋于0的极限3
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1.[㏑(x-π/2)]/tan x 当x趋于π/2时的极限
=lim(x->π/2)1/(x-π/2)/sec²x
=lim(x->π/2)cos²x/(x-π/2)
=lim(x->π/2)2cosx(-sinx)/1
=0
2.
lim(x->0)cotx-1/x
=lim(x->0)xosx/sinx-1/x
=lim(x->0)(xcosx-sinx)/xsinx
=lim(x->0)(xcosx-sinx)/x²
=lim(x->0)(cosx-xsinx-cosx)/2x
=lim(x->0)(-xsinx)/2x
=lim(x->0)(-sinx)/2
=0
3.（x^3+x^2+x+1)^1/3-x
=lim(x->∞)e^ln（x^3+x^2+x+1)^1/3-x
=lim(x->∞)e^[ln（x^3+x^2+x+1)]/(3-x)
=e^lim(x->∞)1/（x^3+x^2+x+1) *(3x²+2x+1)/(-1)
=e^lim(x->∞)-(3x²+2x+1)/（x^3+x^2+x+1)
=e^lim(x->∞)-(6x+2)/（3x^2+2x+1)
=e^lim(x->∞)-(6)/（6x+2)
=e^0
=1
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