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Question Number 140007 by SOMEDAVONG last updated on 03/May/21

$$\underset{x\to 0}{\lim }{(1+\frac{3}{\mathrm{x}})}^{\mathrm{x}}=?$$

Answered by Ankushkumarparcha last updated on 03/May/21

$$Solution:\underset{x\to 0}{\lim }{(1+\frac{3}{x})}^x=\underset{x\to 0}{\lim }{e}^{{\log }_e{(1+\frac{3}{x})}^x}=>\underset{x\to 0}{\lim }{e}^{x{\log }_e(1+\frac{3}{x})}$$$$=\underset{x\to 0}{\lim }{e}^{\frac{{\log }_e(1+\frac{3}{x})}{\frac{1}{x}}}Using{L}^{\prime }HospitalRule.Weget,$$$$=\underset{x\to 0}{\lim }{e}^{\frac{\frac{3}{1+\frac{3}{x}}\cdot \cancel{\frac{-1}{x^2}}}{\cancel{\frac{-1}{x^2}}}}=>\underset{x\to 0}{\lim }{e}^{\frac{3x}{x+3}}$$$$\underset{x\to 0}{\lim }{(1+\frac{3}{x})}^x=1$$

Answered by ajfour last updated on 03/May/21

$$\ln L=\underset{x\to 0}{\lim }\frac{\ln (1+\frac{3}{x})}{\frac{1}{x}}$$$$\ln L=\underset{h\to \mathrm{\infty }}{\lim }\frac{\ln (1+3h)}{h}$$$$usingHopital$$$$=\frac{3}{1+3h}=0$$$$\Rightarrow L=\underset{x\to 0}{\lim }{(1+\frac{3}{\mathrm{x}})}^{\mathrm{x}}=1$$

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