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Na wstepie mamy [0/0], rozklad na iloczyny i skrocenie ulamka. Z definicji Heinego x→1 dla x≠1.
(x²+x-2)/(x⁴-1) = (x²-x+2x-2)/[(x²-1)(x²+1)] =
[x(x-1)+2(x-1)] / [(x-1)(x+1)(x²+1)] = [(x-1)(x+2)]/[(x-1)(x+1)(x²+1)] =
(x+2)/[(x+1)(x²+1)] → (1+2)/(2*2)=3/4
x→1