lim-x-2-x-3-x-1-x-

Question Number 99344 by bemath last updated on 20/Jun/20

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

Answered by abdomsup last updated on 20/Jun/20

let f(x)=(2^x +3^x )^(1/x) ⇒ f(x) =3(1+((2/3))^x )^(1/x) =3 e^((1/x)ln(1+((2/3))^x )) ∼3 e^((1/x)×((2/3))^x ) but lim_(x→+∞) (1/x)((2/3))^x =0 ⇒ lim_(x→+∞) f(x) =3

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

Commented by bemath last updated on 20/Jun/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by abdomsup last updated on 20/Jun/20

$${you}\:{are}\:{welcome} \\ $$

Answered by 4635 last updated on 20/Jun/20

=lim_(x→+∞) e^((1/x)ln (2^x +3^x )) let u=(1/x) ⇒when x⇒+∞ ⇒u⇒0 ⇔lim_(u→0) e^(uln (2^(1/u) +3^(1/u) )) =lim_(u→0) e^(ln 2) ×e^(ln 3) =6 in conclusion lim_(x→+∞) (2^x +3^x )^(1/x) =6

$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{e}^{\frac{\mathrm{1}}{{x}}\mathrm{ln}\:\left(\mathrm{2}^{{x}} +\mathrm{3}^{{x}} \right)} \\ $$$${let}\:{u}=\frac{\mathrm{1}}{{x}}\:\Rightarrow{when}\:{x}\Rightarrow+\infty\:\Rightarrow{u}\Rightarrow\mathrm{0} \\ $$$$\Leftrightarrow\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{{u}\mathrm{ln}\:\left(\mathrm{2}^{\frac{\mathrm{1}}{{u}}} +\mathrm{3}^{\frac{\mathrm{1}}{{u}}} \right)} =\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{\mathrm{ln}\:\mathrm{2}} ×{e}^{\mathrm{ln}\:\mathrm{3}} =\mathrm{6} \\ $$$${in}\:{conclusion}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{2}^{{x}} +\mathrm{3}^{{x}} \right)^{\frac{\mathrm{1}}{{x}}} =\mathrm{6} \\ $$$$ \\ $$

Commented by bobhans last updated on 20/Jun/20

lim_(u→0) e^(u ln(2^(1/u) +3^(1/u) )) = lim_(u→0) e^(ln 2) ×e^(ln 3) ???

$$\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}e}^{\mathrm{u}\:\mathrm{ln}\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{u}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{u}}} \right)} \:=\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{e}^{\mathrm{ln}\:\mathrm{2}} ×\mathrm{e}^{\mathrm{ln}\:\mathrm{3}} \:??? \\ $$

Commented by bobhans last updated on 20/Jun/20

$$\mathrm{your}\:\mathrm{answer}\:\mathrm{not}\:\mathrm{correct} \\ $$

Answered by john santu last updated on 20/Jun/20

lim_(x→∞) (3^x ((2^x /3^x ) +1))^(1/x) = 3 × lim_(x→∞) (1+((2/3))^x )^(1/x) = 3×e^(lim_(x→∞) ln(1+((2/3))^x )^(1/x) ) = 3×e^(lim_(x→∞) ((ln(1+((2/3))^x ))/x)) = 3×1 = 3 ■

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{3}^{\mathrm{x}} \left(\frac{\mathrm{2}^{\mathrm{x}} }{\mathrm{3}^{\mathrm{x}} }\:+\mathrm{1}\right)\right)^{\frac{\mathrm{1}}{\mathrm{x}}} =\:\mathrm{3}\:×\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} \right)^{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$$$=\:\mathrm{3}×\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} \right)^{\frac{\mathrm{1}}{\mathrm{x}}} } \\ $$$$=\:\mathrm{3}×\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} \right)}{\mathrm{x}}} =\:\mathrm{3}×\mathrm{1}\:=\:\mathrm{3}\:\blacksquare \\ $$

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