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Question Number 153404 by liberty last updated on 07/Sep/21

lim_(x→∞) ((8^x +3^x ))^(1/3) −(√(4^x −2^x )) =?

$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\mathrm{8}^{{x}} +\mathrm{3}^{{x}} }\:−\sqrt{\mathrm{4}^{{x}} −\mathrm{2}^{{x}} }\:=? \\ $$

Answered by EDWIN88 last updated on 07/Sep/21

lim_(x→∞) ((8^x +3^x ))^(1/3) −(8^x )^(1/3) +(√4^x )−(√(4^x −2^x )) = L_1 =lim_(x→∞) (((8^x +3^x )−8^x )/( (((8^x +3^x )^2 ))^(1/3) +((8^x (8^x +3^x )))^(1/3) +4^x )) L_1 =lim_(x→∞) (3^x /( ((8^(2x) (1+((3/8))^x )^2 ))^(1/3) +((8^(2x) (1+((3/8))^x )))^(1/3) +4^x )) L_1 =lim_(x→∞) ((((3/4))^x )/( (((1+((3/8))^x )^2 ))^(1/3) +((1+((3/8))^x ))^(1/3) +1)) L_1 =(0/3)=0 L_2 =lim_(x→∞) (√4^x )−(√(4^x −2^x )) L_2 =lim_(x→0) 2^x (1−(√(1−((1/2))^x ))) L_2 =lim_(X→0) ((1−(√(1−X)))/X) = ((1−(1−(1/2)X))/X)=(1/2) Then L=L_1 +L_2 =0+(1/2)=(1/2)

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{3}}]{\mathrm{8}^{{x}} +\mathrm{3}^{{x}} }−\sqrt[{\mathrm{3}}]{\mathrm{8}^{{x}} }\:+\sqrt{\mathrm{4}^{{x}} }−\sqrt{\mathrm{4}^{{x}} −\mathrm{2}^{{x}} }\:= \\ $$$$\:{L}_{\mathrm{1}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{8}^{{x}} +\mathrm{3}^{{x}} \right)−\mathrm{8}^{{x}} }{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{8}^{{x}} +\mathrm{3}^{{x}} \right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{\mathrm{8}^{{x}} \left(\mathrm{8}^{{x}} +\mathrm{3}^{{x}} \right)}+\mathrm{4}^{{x}} } \\ $$$$\:{L}_{\mathrm{1}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}^{{x}} }{\:\sqrt[{\mathrm{3}}]{\mathrm{8}^{\mathrm{2}{x}} \left(\mathrm{1}+\left(\frac{\mathrm{3}}{\mathrm{8}}\right)^{{x}} \right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{\mathrm{8}^{\mathrm{2}{x}} \left(\mathrm{1}+\left(\frac{\mathrm{3}}{\mathrm{8}}\right)^{{x}} \right)}+\mathrm{4}^{{x}} } \\ $$$${L}_{\mathrm{1}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{x}} }{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\left(\frac{\mathrm{3}}{\mathrm{8}}\right)^{{x}} \right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{\mathrm{1}+\left(\frac{\mathrm{3}}{\mathrm{8}}\right)^{{x}} }+\mathrm{1}} \\ $$$${L}_{\mathrm{1}} =\frac{\mathrm{0}}{\mathrm{3}}=\mathrm{0} \\ $$$${L}_{\mathrm{2}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt{\mathrm{4}^{{x}} }−\sqrt{\mathrm{4}^{{x}} −\mathrm{2}^{{x}} }\: \\ $$$${L}_{\mathrm{2}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{2}^{{x}} \:\left(\mathrm{1}−\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} }\right) \\ $$$${L}_{\mathrm{2}} =\underset{{X}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt{\mathrm{1}−{X}}}{{X}}\:=\:\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{X}\right)}{{X}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:{Then}\:{L}={L}_{\mathrm{1}} +{L}_{\mathrm{2}} =\mathrm{0}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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