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Question Number 162827 by saboorhalimi last updated on 01/Jan/22

Answered by Ar Brandon last updated on 01/Jan/22

g(x)=lim_(r→0) ((x+1)^(r+1) −x^(r+1) )^(1/r) lim_(x→∞) ((g(x))/x)=lim_(r→0, x→∞) x^(1+(1/r)−1) ((1+(1/x))^(r+1) −1)^(1/r) =lim_(r→0, x→∞) x^(1/r) (1+((r+1)/x)−1)^(1/r) =lim_(r→0) (r+1)^(1/r) =lim_(r→0) e^((1/r)ln(r+1)) =lim_(r→0) e^((1/r)(r)) =e

$$\mathrm{g}\left({x}\right)=\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\left({x}+\mathrm{1}\right)^{{r}+\mathrm{1}} −{x}^{{r}+\mathrm{1}} \right)^{\frac{\mathrm{1}}{{r}}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{g}\left({x}\right)}{{x}}=\underset{{r}\rightarrow\mathrm{0},\:{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\mathrm{1}+\frac{\mathrm{1}}{{r}}−\mathrm{1}} \left(\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{r}+\mathrm{1}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{{r}}} \\ $$$$=\underset{{r}\rightarrow\mathrm{0},\:{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\frac{\mathrm{1}}{{r}}} \left(\mathrm{1}+\frac{{r}+\mathrm{1}}{{x}}−\mathrm{1}\right)^{\frac{\mathrm{1}}{{r}}} =\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({r}+\mathrm{1}\right)^{\frac{\mathrm{1}}{{r}}} \\ $$$$=\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{e}^{\frac{\mathrm{1}}{{r}}\mathrm{ln}\left({r}+\mathrm{1}\right)} =\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{e}^{\frac{\mathrm{1}}{{r}}\left({r}\right)} ={e} \\ $$

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