about limit of exponential function

Question

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

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

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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

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

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

Hence

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

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