Question Number 140007 by SOMEDAVONG last updated on 03/May/21
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}}\right)^{\mathrm{x}} =? \\ $$
Answered by Ankushkumarparcha last updated on 03/May/21
$${Solution}:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)^{{x}} =\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{\mathrm{log}_{{e}} \left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)^{{x}} } =>\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{{x}\mathrm{log}_{{e}} \left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{\frac{\mathrm{log}_{{e}} \left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)}{\frac{\mathrm{1}}{{x}}}} {Using}\:{L}'{Hospital}\:{Rule}.{We}\:{get}, \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{\frac{\frac{\mathrm{3}}{\mathrm{1}+\frac{\mathrm{3}}{{x}}}\:\centerdot\:\cancel{\frac{−\mathrm{1}}{{x}^{\mathrm{2}} }}}{\cancel{\frac{−\mathrm{1}}{{x}^{\mathrm{2}} }}}} \:=>\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{\frac{\mathrm{3}{x}}{{x}+\mathrm{3}}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)^{{x}} \:=\:\mathrm{1} \\ $$
Answered by ajfour last updated on 03/May/21
$$\mathrm{ln}\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)}{\frac{\mathrm{1}}{{x}}} \\ $$$$\mathrm{ln}\:{L}=\underset{{h}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{3}{h}\right)}{{h}} \\ $$$$\:\:\:\:\:\:{using}\:{Hopital} \\ $$$$\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{1}+\mathrm{3}{h}}=\mathrm{0} \\ $$$$\Rightarrow\:\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}}\right)^{\mathrm{x}} =\:\mathrm{1} \\ $$