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Question Number 31709 by gunawan last updated on 13/Mar/18

$$\mathrm{Given}\mathrm{sequence}\left(y_n\right)\mathrm{with}{y}_1=1,$$$$y_{n+1}=\frac{1}{4}(y_n^3+y_n^2)-1$$$$\mathrm{find}\underset{x\to \mathrm{\infty }}{\lim }{y}_n$$

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