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Question Number 128185 by liberty last updated on 05/Jan/21

 lim_(x→∞) (((x+3)/(x−1)))^(x+3)  =?

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

Answered by john_santu last updated on 05/Jan/21

 lim_(x→∞) (((x−1+4)/(x−1)))^(x+3) =lim_(x→∞) (1+(4/(x−1)))^(x+3)    = lim_(x→∞) (1+(4/(x−1)))^((x−1)+4)    = lim_(h→∞) (1+(4/h))^(h+4)      = lim_(h→∞) (1+(4/h))^h . lim_(h→∞) (1+(4/h))^4    = e^4 .1 = e^4

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}−\mathrm{1}+\mathrm{4}}{{x}−\mathrm{1}}\right)^{{x}+\mathrm{3}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{4}}{{x}−\mathrm{1}}\right)^{{x}+\mathrm{3}} \\ $$$$\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{4}}{{x}−\mathrm{1}}\right)^{\left({x}−\mathrm{1}\right)+\mathrm{4}} \\ $$$$\:=\:\underset{{h}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{4}}{{h}}\right)^{{h}+\mathrm{4}} \:\: \\ $$$$\:=\:\underset{{h}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{4}}{{h}}\right)^{{h}} .\:\underset{{h}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{4}}{{h}}\right)^{\mathrm{4}} \\ $$$$\:=\:{e}^{\mathrm{4}} .\mathrm{1}\:=\:{e}^{\mathrm{4}} \: \\ $$

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