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Find $$\lim_{x\to\infty}x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)$$

I calculated $\lim\limits_{x\to\infty}\left(\frac{x+2}{x+1}\right)^x=e$ but then the limit in question becomes $0\times \infty $ form, and further solution becomes messy.


Please tell a solution without the use of series expansions because I have no knowledge of these.

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  • $\begingroup$ Take $x$ in the denominator. $x=\frac1{\frac1x}$ $\endgroup$ – Aditya Agarwal Sep 25 '15 at 8:38
  • $\begingroup$ And apply LHopital's Rule. $\endgroup$ – Aditya Agarwal Sep 25 '15 at 8:39
  • $\begingroup$ I did the same thing which you are suggesting,but did not reach the solution. $\endgroup$ – Vinod Kumar Punia Sep 25 '15 at 8:41
  • $\begingroup$ Series expansions are useful almost everywhere, they are not complicated. You could maybe read yourself into Taylor series first. CF en.wikipedia.org/wiki/Taylor_series $\endgroup$ – Math-fun Sep 25 '15 at 10:33
  • $\begingroup$ It is not hard to calculate the limit without series by L'Hospital (see below). $\endgroup$ – A.Γ. Sep 25 '15 at 11:42
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\begin{align} \lim_{x\to\infty}x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)&=\lim_{x\to\infty}\frac{e-\left(\frac{x+2}{x+1}\right)^x}{\frac1x}\\ &=\lim_{x\to\infty}\frac{\frac{\left(\frac{1}{x+1}+1\right)^x \left(x-(x+1) (x+2) \log \left(\frac{1}{x+1}+1\right)\right)}{(x+1) (x+2)}}{-\frac1{x^2}}\\ &=-e\lim_{x\to\infty}\left(x-(x+1) (x+2) \log \left(\frac{1}{x+1}+1\right)\right)\\ &=-e\lim_{x\to0}\frac{x-(x+1) (2 x+1) \left(x-\frac{3 x^2}{2}+\frac{7 x^3}{3}-\frac{15 x^4}{4}+o(x^4)\right)}{x^2}\\ &=-e \lim_{x\to 0}(\frac{15 x^4}{2}+\frac{79 x^3}{12}-\frac{x^2}{4}+\frac{x}{6}-\frac{3}{2}+o(x^2))\\ &=\frac32e \end{align}

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Rewriting $\left(\frac{x+2}{x+1}\right)^x=\left(1+\frac{1}{x+1}\right)^x$ with Laurent series expansion rather than calculating out the limit you get: $$\left(\frac{x+2}{x+1}\right)^x=\left(1+\frac{1}{x+1}\right)^x=e - \frac{3e}{2x} +\frac{83e}{24x^2} +\text{some higher order terms}$$ Thus, $$x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)=xe-x\left(e - \frac{3e}{2x} +\frac{83e}{24x^2} +\text{some h.o.t.}\right)=\frac{3e}{2} +\frac{83e}{24x} +\text{some h.o.t.}$$ hence $$ \lim_{x\to\infty}x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)=\frac{3e}{2}$$

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  • $\begingroup$ In the series expansion,did you use binomial theorem or what?How does $e$ come? $\endgroup$ – Vinod Kumar Punia Sep 25 '15 at 8:51
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    $\begingroup$ @VinodKumarPunia, it is Laurent series expansion. $\endgroup$ – Aditya Agarwal Sep 25 '15 at 8:53
  • $\begingroup$ Cant it be solved without Laurent?I have no knowledge of Laurent series. $\endgroup$ – Vinod Kumar Punia Sep 25 '15 at 8:56
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    $\begingroup$ Indeed the crux of the matter is why $\left(1+\frac{1}{x+1}\right)^x=e - \frac{3e}{2x} +\text{some higher order terms}$, and this step might be supplemented by some more details. $\endgroup$ – Did Sep 25 '15 at 9:25
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Without Taylor series: change the variable $t=\frac{1}{x+1}$ to get $$ x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)= \frac{1-t}{t}\left(e-(1+t)^{\frac{1}{t}-1}\right)= \frac{1-t}{1+t}\cdot \frac{e(1+t)-(1+t)^{1/t}}{t}=\\ =\underbrace{\frac{1-t}{1+t}}_{\to 1}e+\underbrace{\frac{1-t}{1+t}}_{\to 1}e\cdot\frac{1-e^{\frac{1}{t}\ln(1+t)-1}}{t}. $$ The first term goes trivially to $1$ as $t\to 0$. Let's calculate the second limit. Denote $s(t)=\frac{1}{t}\ln(1+t)-1$. We have $$ \frac{1-e^{s(t)}}{t}=\frac{1-e^{s(t)}}{s(t)}\cdot\frac{s(t)}{t}= \underbrace{\frac{1-e^{s(t)}}{s(t)}}_{I}\cdot\underbrace{\frac{\ln(1+t)-t}{t^2}}_{II}. $$ The limits $I$ and $II$ are calculated by L'Hospital: \begin{eqnarray} \lim_{t\to 0}s(t)&=&\lim_{t\to 0}\frac{\ln(1+t)}{t}-1=\lim_{t\to 0}\frac{\frac{1}{1+t}}{1}-1=0,\\ \lim I&=&\lim_{s\to 0}\frac{1-e^s}{s}=\lim_{s\to 0}\frac{-e^s}{1}=-1.\\ \lim II&=&\lim_{t\to 0}\frac{\ln(1+t)-t}{t^2}=\lim_{t\to 0}\frac{\frac{1}{1+t}-1}{2t}=\lim_{t\to 0}\frac{-t}{2t(1+t)}=-\frac{1}{2}. \end{eqnarray} Finally, the limit is $e+e\cdot (-1)\cdot\frac{-1}{2}=\frac32 e$.

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