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3.6 pavyzdys. Apskaičiuokime ribą√x 2 + x − x.Sprendimas.limx→∞√lim x 2 ( √ x+ x − x = (∞ − ∞) = lim2 + x − x)( √ x 2 + x + x)√x→∞x→∞ x 2 + x + x= limx→∞x 2 + x − x 2√x 2 + x + x = lim x→∞x√x 2 + x + x= [skaitiklį ir vardiklį dalijame iš x]= limx→∞xx√ x 2x 2 + x x 2 + x x= limx→∞1√ = 11 + 1 x + 1 2 . Atsakymas. 1 2 .3.1 užduotis savarankiškam darbui.Užd. Riba Užd. Ribax3.1.1 lim3 −3x−23.1.11 limx→∞x+x 2 x→∞3.1.2 limx→∞3.1.3 limx→∞3.1.4 limx→∞3.1.5 limx→∞3.1.6 limx→∞x 4 −12x 4 −x 2 −13.1.12 limx→∞(1+x) 3 −(1+3x)x+x 5 3.1.13 lim3x 3 −5x+1x→∞4−x 2 −2x 3 3.1.14 limx→∞4−x 42+x+3x 2 3.1.15 limx→∞(2x+1) 4 −(x−1) 4(2x+1) 4 +(x−1) 4 3.1.16 lim4x 2 −(x+4) 23.1.7 lim3.1.17 limx→∞(x−1)(2x+3)(2−5x)x→∞3√3.1.8 lim x−6+23.1.18 limx→+∞x 3 +8√ √3.1.9 lim x+19−2 x+13.1.19 limx 2 −9x→+∞3.1.10 limx→+∞(√ 3x + 5 −√ 3x − 7)3.1.20 limx→+∞x 2 −3x−2x+x 2x 4 +12x 4 −x 2 +1(1+x) 2 −(1+3x)x+x 53x 3 +5x+14−x 2 −2x 34+x 32+x+3x 2(2x+1) 2 −(x−1) 2(2x+1) 2 +(x−1) 2x→∞4x 2 −(x+4) 2(x+1)(2x+3)(2−x)3√ x−6+2x→+∞x 2 −1√ √ x+19+2 x+1x→+∞x 2 −1(√ x + 3 −√ x − 2)35