高等数学试题绪悄趋\x0d1.计算lim(1/x2-cot2x) X→0 2 lim(1-x)tanπx/2x→1

= lim (1/x^2-1/tan2x)

= lim (tan2x-x^2)/(x^2tan2x) = lim (tan2x-x^2)/(2x^3) (0/0)

= lim [(sec2x)^2-x]/(3x^2) = ∞

若是 lim [1/x^2-(cotx) ^2] 则该极限

= lim [1/x^2-(cosx)^2/(sinx)^2]

= lim [(sinx)^2-x^2(cosx)^2]/[x^2(sinx)^2]

= lim [(sinx)^2-x^2(cosx)^2]/x^4 (0/0)

= lim [sin2x-2x(cosx)^2+x^2sin2x]/(4x^3) (0/0)

= lim [cos2x-(cosx)^2+2xsin2x+x^2cos2x]/(6x^2) (0/0)

= lim [sin2x+6xcos2x-2x^2sin2x]/(12x) (0/0)

= lim [8cos2x-16xsin2x-4x^2cos2x]/12 = 2/3.

2. lim (1-x)tan(πx/2) = lim (1-x)/cot(πx/2) (0/0)

= lim -1/{(-π/2)[csc(πx/2)]^2} = 2/π.