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Question Number 1635 by 123456 last updated on 28/Aug/15
$$\mathrm{lets}\:{x}>\mathrm{0},\:\mathrm{and}\:\mathrm{take}\:\mathrm{the}\:\mathrm{sequence}\:{a} \\ $$ $${a}_{\mathrm{0}} =\sqrt{{x}} \\ $$ $${a}_{{n}+\mathrm{1}} =\sqrt{{x}+{a}_{{n}} } \\ $$ $$\mathrm{i}.\mathrm{proof}\:\mathrm{that}\:\mathrm{0}\leqslant{a}_{{n}} \leqslant{a}_{{n}+\mathrm{1}} \\ $$ $$\mathrm{ii}.\mathrm{proof}\:\mathrm{that}\:\exists\mathrm{M}\:\mathrm{such}\:\mathrm{that}\:{a}_{{n}} \leqslant\mathrm{M} \\ $$ $$\mathrm{iii}.\mathrm{using}\:\mathrm{i}\:\mathrm{and}\:\mathrm{ii}\:\mathrm{proof}\:\mathrm{that}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} \:\mathrm{exist} \\ $$ $$\mathrm{iv}.\mathrm{compute}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} \\ $$