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
$${let}\:{f}\left({x}\right)=\left(\mathrm{2}^{{x}} \:+\mathrm{3}^{{x}} \right)^{\frac{\mathrm{1}}{{x}}} \:\Rightarrow \\ $$$${f}\left({x}\right)\:=\mathrm{3}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} \right)^{\frac{\mathrm{1}}{{x}}} \\ $$$$=\mathrm{3}\:{e}^{\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} \right)} \:\sim\mathrm{3}\:{e}^{\frac{\mathrm{1}}{{x}}×\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} } \\ $$$${but}\:{lim}_{{x}\rightarrow+\infty} \frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} \:=\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)\:=\mathrm{3} \\ $$
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
$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{e}^{\frac{\mathrm{1}}{{x}}\mathrm{ln}\:\left(\mathrm{2}^{{x}} +\mathrm{3}^{{x}} \right)} \\ $$$${let}\:{u}=\frac{\mathrm{1}}{{x}}\:\Rightarrow{when}\:{x}\Rightarrow+\infty\:\Rightarrow{u}\Rightarrow\mathrm{0} \\ $$$$\Leftrightarrow\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{{u}\mathrm{ln}\:\left(\mathrm{2}^{\frac{\mathrm{1}}{{u}}} +\mathrm{3}^{\frac{\mathrm{1}}{{u}}} \right)} =\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{\mathrm{ln}\:\mathrm{2}} ×{e}^{\mathrm{ln}\:\mathrm{3}} =\mathrm{6} \\ $$$${in}\:{conclusion}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{2}^{{x}} +\mathrm{3}^{{x}} \right)^{\frac{\mathrm{1}}{{x}}} =\mathrm{6} \\ $$$$ \\ $$
Commented by bobhans last updated on 20/Jun/20
$$\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}e}^{\mathrm{u}\:\mathrm{ln}\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{u}}} +\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{u}}} \right)} \:=\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{e}^{\mathrm{ln}\:\mathrm{2}} ×\mathrm{e}^{\mathrm{ln}\:\mathrm{3}} \:??? \\ $$
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
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{3}^{\mathrm{x}} \left(\frac{\mathrm{2}^{\mathrm{x}} }{\mathrm{3}^{\mathrm{x}} }\:+\mathrm{1}\right)\right)^{\frac{\mathrm{1}}{\mathrm{x}}} =\:\mathrm{3}\:×\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} \right)^{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$$$=\:\mathrm{3}×\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} \right)^{\frac{\mathrm{1}}{\mathrm{x}}} } \\ $$$$=\:\mathrm{3}×\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} \right)}{\mathrm{x}}} =\:\mathrm{3}×\mathrm{1}\:=\:\mathrm{3}\:\blacksquare \\ $$