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Maybe the answer is obvious. I'm sorry for this

I know for all $x \in \mathbb{R}$ that

$$ \lim_\limits{n \to \infty}\left(1 + \frac{x}{n} \right)^{n} = \exp(x). $$

Now suppose I have a sequence $\{x_{n}\}_{n \in \mathbb{N}}$ such that

$$ \lim_\limits{n \to \infty} x_{n} = x \in \mathbb{R}. $$

Can I also conclude that

$$ \lim_\limits{n \to \infty}\left(1 + \frac{x_{n}}{n} \right)^{n} = \exp(x)? $$


marked as duplicate by RRL real-analysis Sep 11 '18 at 16:58

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  • $\begingroup$ I think yes but not sure $\endgroup$ – Cloud JR Sep 11 '18 at 15:56
  • $\begingroup$ $\lim_{n\to +\infty} x_n=x$ means that for any $\varepsilon > 0$ there is some $N$ ensuring $|x_n-x|\leq \varepsilon$ for any $n\geq N$. In particular, for any $n\geq N$ we have that $\liminf\left(1+x_n/n\right)^n\geq \exp(x-\varepsilon)$ and $\limsup\left(1+x_n/n\right)^n\leq \exp(x+\varepsilon)$. Now we may exploit the fact that $\varepsilon$ is arbitrarly small. $\endgroup$ – Jack D'Aurizio Sep 11 '18 at 15:57

We have that $\forall \epsilon>0$ eventually

$$\left(1 + \frac{x-\epsilon}{n} \right)^{n}\le \left(1 + \frac{x_{n}}{n} \right)^{n}\le \left(1 + \frac{x+\epsilon}{n} \right)^{n}$$

and therefore by squeeze theorem

$$e^{x-\epsilon} \le \liminf_{n \to \infty}\left(1 + \frac{x_{n}}{n} \right)^{n}\le \limsup_{n \to \infty}\left(1 + \frac{x_{n}}{n} \right)^{n}\le e^{x+\epsilon} $$

and taking $\epsilon \to 0$ the result follows.

  • $\begingroup$ The inequality holds only for $n \geq N$ for some $N$ depending on $\varepsilon$ right? $\endgroup$ – gregor Sep 11 '18 at 16:11
  • $\begingroup$ Yes of course and in the limit for $\epsilon \to 0$ we obtain the result. $\endgroup$ – gimusi Sep 11 '18 at 16:57
  • $\begingroup$ We do not know, a priori, that $\lim_{n\to\infty}\left(1+\frac xn\right)^n$ exists. Use $\liminf$ and $\limsup$. $\endgroup$ – Mark Viola Sep 11 '18 at 16:57
  • $\begingroup$ @MarkViola yes of course you are right! I fix that. Thanks $\endgroup$ – gimusi Sep 11 '18 at 17:04

So, can one conclude that $$n\ln\left(1+\frac{x_n}n\right)\to x?\tag{*}$$ Note that $$n\ln\left(1+\frac{x_n}n\right)=x_n+O\left(\frac{x_n^2}{n}\right).$$ But $x_n\to x$, so $O(x_n^2/n)=O(1/n)$. Thus (*) holds, and the answer is yes.


By mean value theorem on $f(x)=\ln(\alpha+x)$ in $[x,x+\alpha]$ one may prove $$\dfrac{x}{\alpha+x}<\ln(\alpha+x)-\ln(x)<1$$ this show $$\lim_{x\to\infty}x\ln\left(1+\dfrac{\alpha}{x}\right)=1$$ which gives the result.


$n\log (1+x_n/n)= x_n\frac{ \log (1+x_n/n)- \log (1)}{x_n/n}=$

$x_n\log '(y_n)=x_n(1/y_n)$, where $1< y_n <1+x_n/n$.

$\lim_{n \rightarrow \infty} y_n = 1$.


$\lim_{n \rightarrow \infty} [x_n (1/y_n)] =$

$[\lim_{n \rightarrow \infty}(x_n)][\lim_{ n \rightarrow \infty}(1/y_n)]= x. $


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