Question Number 156677 by saly last updated on 14/Oct/21
Commented by cortano last updated on 14/Oct/21
$$\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\sqrt[{\mathrm{n}}]{\mathrm{a}}+\sqrt[{\mathrm{n}}]{\mathrm{b}}}{\mathrm{2}}\right)^{\mathrm{n}} ? \\ $$
Commented by saly last updated on 14/Oct/21
$${whath}? \\ $$
Commented by john_santu last updated on 14/Oct/21
$${if}\:{A}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\sqrt[{{n}}]{{a}}+\sqrt[{{n}}]{{b}}}{\mathrm{2}}\right)^{{n}} =\left(\frac{\sqrt[{{n}}]{{a}}+\sqrt[{{b}}]{{n}}}{\mathrm{2}}\right)^{{n}} ×\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}\right) \\ $$$$=\:\left(\frac{\sqrt[{{n}}]{{a}}\:+\sqrt[{{n}}]{{b}}}{\mathrm{2}}\:\right)^{{n}} \\ $$
Answered by puissant last updated on 14/Oct/21
$${A}=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\sqrt[{{n}}]{{a}}+\sqrt[{{n}}]{{b}}}{\mathrm{2}}\right)^{{n}} \:;\:\:{x}=\frac{\mathrm{1}}{{n}}\:\rightarrow{n}=\frac{\mathrm{1}}{{x}} \\ $$$${A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{a}^{{x}} +{b}^{{x}} }{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}}} =\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}+{xlna}+\mathrm{1}+{xlnb}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}{ln}\left({ab}\right)\right)^{\frac{\mathrm{1}}{{x}}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{\frac{{ln}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}{lnab}\right)}{{x}}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{\frac{\frac{{x}}{\mathrm{2}}{lnab}}{{x}}} \:\:=\:\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{lnab}} \:\:=\:\:\sqrt{{ab}}… \\ $$$$\:\:\:\:\:\:\therefore\because\:\:{A}\:=\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\sqrt[{{n}}]{{a}}+\sqrt[{{n}}]{{b}}}{\mathrm{2}}\right)^{{n}} =\:\sqrt{{ab}}…. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…………\mathscr{L}{e}\:{puissant}………… \\ $$