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Question Number 126376 by benjo_mathlover last updated on 20/Dec/20

$$\underset{x\to 0}{\lim }{(1+3x)}^{-\frac{5}{x}}?$$

Answered by liberty last updated on 20/Dec/20

$$\underset{x\to 0}{\lim }{(1+3x)}^{-\frac{5}{x}}=e^{\underset{x\to 0}{\lim }(1+3x-1)(-\frac{5}{x})}$$$$=e^{\underset{x\to 0}{\lim }-\left(\frac{15x}{x}\right)}=e^{-15}$$

Answered by Dwaipayan Shikari last updated on 20/Dec/20

$$\underset{x\to 0}{\lim }{(1+3x)}^{-\frac{5}{x}}=\underset{x\to 0}{\lim }{(1+3x)}^{\frac{-15}{3x}}=\frac{1}{e^{15}}$$

Answered by mathmax by abdo last updated on 20/Dec/20

$$\mathrm{f}\left(\mathrm{x}\right)={(1+3\mathrm{x})}^{\frac{-5}{\mathrm{x}}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{-\frac{5}{\mathrm{x}}\ln (1+3\mathrm{x})}$$$$\mathrm{we}\mathrm{have}\mathrm{for}\mathrm{x}\sim 0\ln (1+3\mathrm{x})\sim 3\mathrm{x}\Rightarrow -\frac{5}{\mathrm{x}}\ln (1+3\mathrm{x})\sim -\frac{5}{\mathrm{x}}\left(3\mathrm{x}\right)=-15\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{-15}$$

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