obliczyć granicę łukasz:

	n+3
(		)n2+2n
	n

i jeszcze jedno

	n+3
(		)n+5
	n−2

nie potrafie tego rozwiązać...

Basia: Liczę

Basia: (ⁿ⁺³_n)ⁿ²⁺²ⁿ = (1+³_n)ⁿ²*(1+³_n)²ⁿ=

	1		1
(1 +		)n2*(1 +		)2n
	n3		n3

podstawiamy t = ⁿ₃ n = 3t jeżeli n→+∞ to t→+∞ mamy (1+¹_t)^9t2*(1+¹_t)^6t = [ [ (1+¹_t)^t ]^9t*[ (1+¹_t)^t ]⁶ → e^9t*e⁶ = e³ⁿ*e⁶ → (+∞)*e⁶ = +∞ −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− (ⁿ⁺³_n−2)ⁿ⁺⁵ = (ⁿ⁻²⁺⁵_n−2)ⁿ⁺⁵ = (1 + ⁵_n−2)ⁿ⁺⁵ =

	1
(1 +		)n+5
	n−25

podstawiamy t = ⁿ⁻²₅ jeżeli n→+∞ to t→+∞ 5t = n−2 n = 5t+2 (1+¹_t)^5t+7 = (1+¹_t)^5t*(1+¹_t)⁷ = [ (1+¹_t)^t ]⁵*(1+¹_t)⁷ → e⁵*(1+0)⁷ = e⁵