Question-125204

Question Number 125204 by bemath last updated on 09/Dec/20

Answered by liberty last updated on 09/Dec/20

lim_(x→∞) (3^(4x) )^(1/(2x)) (1+(1/3^(3x) ))^(1/(2x)) = 9 × lim_(x→∞) (1+(1/3^(3x) ))^(1/(2x)) = 9 × e^(lim_(x→∞) (1+(1/3^(2x) )−1).(1/(2x))) = 9 × e^(lim_(x→∞) ((1/(2x. 3^(2x) )))) = 9×e^0 = 9

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{3}^{\mathrm{4}{x}} \right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}{x}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} =\: \\ $$$$\:\mathrm{9}\:×\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}{x}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \:=\:\mathrm{9}\:×\:{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}{x}} }−\mathrm{1}\right).\frac{\mathrm{1}}{\mathrm{2}{x}}} \\ $$$$\:=\:\mathrm{9}\:×\:{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{2}{x}.\:\mathrm{3}^{\mathrm{2}{x}} }\right)} \:=\:\mathrm{9}×{e}^{\mathrm{0}} =\:\mathrm{9} \\ $$

Answered by Dwaipayan Shikari last updated on 09/Dec/20

lim_(x→∞) 3^2 (1+(1/3^(3x) ))^(1/(2x)) =lim_(x→∞) 3^2 (1+(1/9^x ))^(1/(2x)) =3^2 (1+(1/(9^x .2x))). =9

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}3}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}{x}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}3}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{9}^{{x}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} =\mathrm{3}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{9}^{{x}} .\mathrm{2}{x}}\right). \\ $$$$=\mathrm{9} \\ $$

Answered by mathmax by abdo last updated on 09/Dec/20

let f(x)=(3^x +3^(4x) )^(1/(2x)) ⇒f(x)=(3^(4x) )^(1/(2x)) {1+(1/3^x )}^(1/(2x)) =9 e^((1/(2x))ln(1+(1/3^x ))) but ln(1+(1/3^x ))∼(1/3^x )(x→+∞) ⇒ e^((1/(2x))ln(1+(1/3^x ))) ∼e^(1/(2x.3^x )) ⇒f(x)∼ 9 e^(1/(2x3^x )) ⇒lim_(x→+∞) f(x)=9

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{3}^{\mathrm{x}} \:+\mathrm{3}^{\mathrm{4x}} \right)^{\frac{\mathrm{1}}{\mathrm{2x}}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{3}^{\mathrm{4x}} \right)^{\frac{\mathrm{1}}{\mathrm{2x}}} \left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{x}} }\right\}^{\frac{\mathrm{1}}{\mathrm{2x}}} \\ $$$$=\mathrm{9}\:\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2x}}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{x}} }\right)} \:\:\mathrm{but}\:\:\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{x}} }\right)\sim\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{x}} }\left(\mathrm{x}\rightarrow+\infty\right)\:\Rightarrow \\ $$$$\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2x}}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{x}} }\right)} \sim\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2x}.\mathrm{3}^{\mathrm{x}} }} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\:\mathrm{9}\:\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2×3}^{\mathrm{x}} }} \:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{9} \\ $$

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