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Question Number 74782 by naka3546 last updated on 30/Nov/19

Commented by abdomathmax last updated on 01/Dec/19

let f(x)={((1/8))^x +((1/2))^x }^(1/x^2 ) ⇒ ln(f(x))=(1/x^2 )ln{ ((1/8))^x +((1/2))^x } =(1/x)ln((1/8)) +(1/x^2 )ln{ 1+4^x } =((−3ln(2))/x)+((2ln(2))/x)+ln{1+4^(−x) } ⇒f(x)∼−((ln(2))/x) +4^(−x) →0(x→+∞) ⇒ lim_(x→+∞) f(x)=1

$${let}\:{f}\left({x}\right)=\left\{\left(\frac{\mathrm{1}}{\mathrm{8}}\right)^{{x}} \:+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} \right\}^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:\Rightarrow \\ $$$${ln}\left({f}\left({x}\right)\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left\{\:\left(\frac{\mathrm{1}}{\mathrm{8}}\right)^{{x}} \:+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} \right\} \\ $$$$=\frac{\mathrm{1}}{{x}}{ln}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left\{\:\mathrm{1}+\mathrm{4}^{{x}} \right\} \\ $$$$=\frac{−\mathrm{3}{ln}\left(\mathrm{2}\right)}{{x}}+\frac{\mathrm{2}{ln}\left(\mathrm{2}\right)}{{x}}+{ln}\left\{\mathrm{1}+\mathrm{4}^{−{x}} \right\} \\ $$$$\Rightarrow{f}\left({x}\right)\sim−\frac{{ln}\left(\mathrm{2}\right)}{{x}}\:+\mathrm{4}^{−{x}} \:\:\rightarrow\mathrm{0}\left({x}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} \:{f}\left({x}\right)=\mathrm{1} \\ $$

Commented by abdomathmax last updated on 24/Dec/19

ln(f(x))∼−((ln(2))/x) +4^(−x) (x→+∞)

$${ln}\left({f}\left({x}\right)\right)\sim−\frac{{ln}\left(\mathrm{2}\right)}{{x}}\:+\mathrm{4}^{−{x}} \:\:\:\left({x}\rightarrow+\infty\right) \\ $$

Answered by MJS last updated on 30/Nov/19

((1/2^(3x) )+(1/2^x ))^(1/x^2 ) =((1/8^x )(1+4^x ))^(1/x^2 ) =(((1+4^x ))^(1/x^2 ) /(8)^(1/x) ) lim_(x→∞) (1/(8)^(1/x) )=1 lim_(x→∞) ((1+4^x ))^(1/x^2 ) ∼lim_(x→∞) (4)^(1/x) =1 ⇒ answer is (C)

$$\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}{x}} }+\frac{\mathrm{1}}{\mathrm{2}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} =\left(\frac{\mathrm{1}}{\mathrm{8}^{{x}} }\left(\mathrm{1}+\mathrm{4}^{{x}} \right)\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} =\frac{\sqrt[{{x}^{\mathrm{2}} }]{\mathrm{1}+\mathrm{4}^{{x}} }}{\sqrt[{{x}}]{\mathrm{8}}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\sqrt[{{x}}]{\mathrm{8}}}=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{{x}^{\mathrm{2}} }]{\mathrm{1}+\mathrm{4}^{{x}} }\sim\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{{x}}]{\mathrm{4}}\:=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\left(\mathrm{C}\right) \\ $$

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