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

  lim_(x→0)  (1+3x)^(−(5/x))  ?

$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\mathrm{3}{x}\right)^{−\frac{\mathrm{5}}{{x}}} \:? \\ $$

Answered by liberty last updated on 20/Dec/20

 lim_(x→0) (1+3x)^(−(5/x)) = e^(lim_(x→0) (1+3x−1)(−(5/x)))    = e^(lim_(x→0) −(((15x)/x))) = e^(−15)

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{3}{x}\right)^{−\frac{\mathrm{5}}{{x}}} =\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{3}{x}−\mathrm{1}\right)\left(−\frac{\mathrm{5}}{{x}}\right)} \\ $$$$\:=\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\left(\frac{\mathrm{15}{x}}{{x}}\right)} =\:{e}^{−\mathrm{15}} \\ $$

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

lim_(x→0) (1+3x)^(−(5/x)) =lim_(x→0) (1+3x)^((−15)/(3x)) =(1/e^(15) )

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

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

f(x)=(1+3x)^((−5)/x)   ⇒f(x)=e^(−(5/x)ln(1+3x))   we have for x ∼0     ln(1+3x)∼3x ⇒−(5/x)ln(1+3x)∼−(5/x)(3x)=−15 ⇒  lim_(x→0) f(x)=e^(−15)

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

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