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Question-64785




Question Number 64785 by Rio Michael last updated on 21/Jul/19
Answered by LPM last updated on 21/Jul/19
1) x_n ≤ 2 ,∀ n≥1  2) x_n ≤x_(n+1) , ∀ n≥1   ⇒ x_n  cov  let x_n → L ⇒ L^2 −2L=0                        ⇒ L = 0 or L =2                        ⇒ L = 2 (x_n >0)    log_e x_n →log_e 2
$$\left.\mathrm{1}\right)\:\mathrm{x}_{\mathrm{n}} \leqslant\:\mathrm{2}\:,\forall\:\mathrm{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{x}_{\mathrm{n}} \leqslant\mathrm{x}_{\mathrm{n}+\mathrm{1}} ,\:\forall\:\mathrm{n}\geqslant\mathrm{1} \\ $$$$\:\Rightarrow\:\mathrm{x}_{\mathrm{n}} \:\mathrm{cov} \\ $$$$\mathrm{let}\:\mathrm{x}_{\mathrm{n}} \rightarrow\:\mathrm{L}\:\Rightarrow\:\mathrm{L}^{\mathrm{2}} −\mathrm{2L}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{L}\:=\:\mathrm{0}\:\mathrm{or}\:\mathrm{L}\:=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{L}\:=\:\mathrm{2}\:\left(\mathrm{x}_{\mathrm{n}} >\mathrm{0}\right) \\ $$$$\:\:\mathrm{log}_{\mathrm{e}} \mathrm{x}_{\mathrm{n}} \rightarrow\mathrm{log}_{\mathrm{e}} \mathrm{2} \\ $$

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