# Limit of $x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)$ when $x\to+\infty$

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.

• Take $x$ in the denominator. $x=\frac1{\frac1x}$ – Aditya Agarwal Sep 25 '15 at 8:38
• And apply LHopital's Rule. – Aditya Agarwal Sep 25 '15 at 8:39
• I did the same thing which you are suggesting,but did not reach the solution. – Vinod Kumar Punia Sep 25 '15 at 8:41
• 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 – Math-fun Sep 25 '15 at 10:33
• It is not hard to calculate the limit without series by L'Hospital (see below). – A.Γ. Sep 25 '15 at 11:42

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 $$\lim_{x\to\infty}x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)=\frac{3e}{2}$$
• In the series expansion,did you use binomial theorem or what?How does $e$ come? – Vinod Kumar Punia Sep 25 '15 at 8:51
• 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. – Did Sep 25 '15 at 9:25
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 $e+e\cdot (-1)\cdot\frac{-1}{2}=\frac32 e$.