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lim-x-0-1-3-x-x-




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

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