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l-n-im-n-1-n-n-1-n-

Question Number 70360 by Hassen_Timol last updated on 03/Oct/19 $$\underset{{n}\rightarrow+\infty} {\mathrm{l}im}\:\:\:\:\frac{\sqrt{{n}\:+\:\mathrm{1}\:}−\:{n}}{\:\sqrt{{n}\:+\:\mathrm{1}}\:+\:{n}}\:\:=\:\:? \\ $$ Answered by mind is power last updated on 03/Oct/19 $$\frac{\sqrt{{n}+\mathrm{1}}−{n}}{\:\sqrt{{n}+\mathrm{1}}+{n}}=\frac{{n}\left(\sqrt{\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} }}−\mathrm{1}\right)}{{n}\left(\sqrt{\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} }}+\mathrm{1}\right)}=\frac{\sqrt{\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}}…

Question-135838

Question Number 135838 by abdurehime last updated on 16/Mar/21 Answered by liberty last updated on 16/Mar/21 $${f}\left({x}\right)=\:\frac{{x}}{{x}^{\mathrm{2}} +{k}}\:\Rightarrow\mathrm{ln}\:{y}\:=\:\mathrm{ln}\:{x}−\mathrm{ln}\:\left({x}^{\mathrm{2}} +{k}\right) \\ $$$$\:\frac{{y}'}{{y}}\:=\:\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +{k}} \\ $$$$\frac{{y}'}{{y}}\:=\:\frac{{k}−{x}^{\mathrm{2}} }{{x}\left({x}^{\mathrm{2}}…