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Question Number 156677 by saly last updated on 14/Oct/21

Commented by cortano last updated on 14/Oct/21

lim_(n→∞) ((((a)^(1/n) +(b)^(1/n) )/2))^n ?

$$\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\sqrt[{\mathrm{n}}]{\mathrm{a}}+\sqrt[{\mathrm{n}}]{\mathrm{b}}}{\mathrm{2}}\right)^{\mathrm{n}} ? \\ $$

Commented by saly last updated on 14/Oct/21

whath?

$${whath}? \\ $$

Commented by john_santu last updated on 14/Oct/21

if A=lim_(x→∞) ((((a)^(1/n) +(b)^(1/n) )/2))^n =((((a)^(1/n) +(n)^(1/b) )/2))^n ×lim_(x→∞) (1) = ((((a)^(1/n) +(b)^(1/n) )/2) )^n

$${if}\:{A}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\sqrt[{{n}}]{{a}}+\sqrt[{{n}}]{{b}}}{\mathrm{2}}\right)^{{n}} =\left(\frac{\sqrt[{{n}}]{{a}}+\sqrt[{{b}}]{{n}}}{\mathrm{2}}\right)^{{n}} ×\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}\right) \\ $$$$=\:\left(\frac{\sqrt[{{n}}]{{a}}\:+\sqrt[{{n}}]{{b}}}{\mathrm{2}}\:\right)^{{n}} \\ $$

Answered by puissant last updated on 14/Oct/21

A=lim_(n→+∞) ((((a)^(1/n) +(b)^(1/n) )/2))^n ; x=(1/n) →n=(1/x) A=lim_(x→0) (((a^x +b^x )/2))^(1/x) = lim_(x→0) (((1+xlna+1+xlnb)/2))^(1/x) =lim_(x→0) (1+(x/2)ln(ab))^(1/x) =lim_(x→0) e^((ln(1+(x/2)lnab))/x) = lim_(x→0) e^(((x/2)lnab)/x) = e^((1/2)lnab) = (√(ab))... ∴∵ A = lim_(n→+∞) ((((a)^(1/n) +(b)^(1/n) )/2))^n = (√(ab)).... ............Le puissant............

$${A}=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\sqrt[{{n}}]{{a}}+\sqrt[{{n}}]{{b}}}{\mathrm{2}}\right)^{{n}} \:;\:\:{x}=\frac{\mathrm{1}}{{n}}\:\rightarrow{n}=\frac{\mathrm{1}}{{x}} \\ $$$${A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{a}^{{x}} +{b}^{{x}} }{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}}} =\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}+{xlna}+\mathrm{1}+{xlnb}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}{ln}\left({ab}\right)\right)^{\frac{\mathrm{1}}{{x}}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{\frac{{ln}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}{lnab}\right)}{{x}}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{\frac{\frac{{x}}{\mathrm{2}}{lnab}}{{x}}} \:\:=\:\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{lnab}} \:\:=\:\:\sqrt{{ab}}... \\ $$$$\:\:\:\:\:\:\therefore\because\:\:{A}\:=\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\sqrt[{{n}}]{{a}}+\sqrt[{{n}}]{{b}}}{\mathrm{2}}\right)^{{n}} =\:\sqrt{{ab}}.... \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:............\mathscr{L}{e}\:{puissant}............ \\ $$

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