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My solution to this problem is as follows:

Let $\epsilon_n$ be a sequence of numbers that converge to $0$. Let $c = \epsilon_n x$, where $x \in \Re$. Then, using the limit definition of $e^x$ we can write:$$\lim_{n \to \infty}(1 + \frac {c}{n})^n = e^c$$ As $n \to \infty$, $c \to 0$ and $e^c \to 1$. $$\therefore \lim_{n \to \infty}(1 + \epsilon_n\frac {x}{n})^n = 1$$

Is this solution valid? I'm not sure if my use of the limit definition of $e^x$ is correct here.

David Choi
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    The formula for $e^c$ in principle is for fixed $c$. Your $c$ depends on $n$. – Gary Jun 18 '21 at 07:13
  • Why not writng $\left(1+\epsilon_n\frac{x}{n}\right)^n=\exp\left[n\ln\left(1+\epsilon_n\frac{x}{n}\right)\right]$ then using Taylor's formula at $1$ for the $\ln$? – Nicolas Jun 18 '21 at 07:15
  • I don’t think that answer is valid. You are taking $c\epsilon_n$ as a constant, and letting it approach zero after letting $n\to\infty.$ But see the Lemma in this answer, particularly the firs case of $0$: https://math.stackexchange.com/a/1451245/7933 – Thomas Andrews Jun 18 '21 at 07:15

4 Answers4

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I guess there are a lot of ways to solve it. But your solution seems lack of some rigour, I try to improve it a little bit in the following

For example, because the sequence $\epsilon_n$ converges to $0$, for all $c>0$ we can always find an $N$ such that $|\epsilon_n x|<c $ for all $n>N$. And that implies that: $$\limsup_n (1+\frac{x\epsilon}{n})^n \le \limsup_n (1+\frac{c}{n})^n =e^c $$ for all $c>0$, hence $$\limsup_n (1+\frac{x\epsilon_n}{n})^n \le 1$$. Similarly, you can also show that: $$\liminf_n (1+\frac{x\epsilon}{n})^n \ge \liminf_n (1-\frac{c}{n})^n =e^{-c} $$ for all $c>0$, Hence $$\liminf_n (1+\frac{x\epsilon_n}{n})^n \ge 1$$ So, from two observations above, we can conclude that: $$\lim_n (1+\frac{x\epsilon_n}{n})^n =1$$
Paresseux Nguyen
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  • If you use \left( and \right) you get better parentheses:$$\left(1+\frac cn\right)^n$$ – Thomas Andrews Jun 18 '21 at 07:20
  • Thank you for your recommendations. The $\le 0$ is just a typo. – Paresseux Nguyen Jun 18 '21 at 07:21
  • For $n$ big enough, you see that $ x\epsilon_n<C$ – Paresseux Nguyen Jun 18 '21 at 09:19
  • in the liminf, it was a typo, the inequality should be the other way. I corrected it. – Paresseux Nguyen Jun 18 '21 at 09:20
  • Sorry, I want to clarify my question. My question is why do you need to use the concept of limit superior and limit inferior in the first place. Since, for all $c>0$, we can always find an N such that $|\epsilon_n x| < c$ for all $n>N$, we can write: $$(1 - \frac {c}{n})^n < (1 + \frac {\epsilon_n x}{n})^n < (1 + \frac {c}{n})^n$$ If we take limit of this inequality as n approaches $\infty$, we get $$e^{-c} < \lim\limits_{n\rightarrow\infty}\left(1+\epsilon_n \frac{x}{n}\right)^n < e^c$$ – David Choi Jun 18 '21 at 10:55
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    Okay. Because, you don't know if $\lim (1+ \frac{\epsilon_nx}{n})^n$ exists, you can't use that expression into your calculations. However, this is not the case with limsup and liminf because they always exist. – Paresseux Nguyen Jun 18 '21 at 10:58
  • One more question, in the following inequalities: $$\limsup_n (1+\frac{x\epsilon_n}{n})^n \le \limsup_n (1+\frac{c}{n})^n =e^c, \ \liminf_n (1+\frac{x\epsilon_n}{n})^n \geq \liminf_n (1-\frac{c}{n})^n =e^{-c}$$ can't we make the inequalities stricter so that its just less than or greater than? As from the inequality $|\epsilon_n x| < c$ we can get the inequality $$(1 + \frac {\epsilon_n x}{n})^n < (1 + \frac {c}{n})^n$$ so can't we apply the limit superior to this inequality? (Similar case for limit inferior) – David Choi Jun 19 '21 at 08:55
  • From the inequality $a_n < b_n$, the best you can have is $\limsup a_n \le \limsup b_n$ ( not the strict inequality), so I guess what you want does not matter in the end. – Paresseux Nguyen Jun 19 '21 at 11:05
  • By all means, to conclude, you don't care if it is the strict inequality or not, so why should we be concerned by that? – Paresseux Nguyen Jun 19 '21 at 11:06
  • I just wanted to ask to deepen my understanding, as I thought we could have a stricter inequality. – David Choi Jun 20 '21 at 00:45
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Without loss take $x \ge 0$. Break $\epsilon_n$ to a positive and/or a negative subsequence.

Considering the subsequence $\epsilon_{n_k} \ge 0$, using $e^y \ge 1+y$ when $y \ge 0$:

\begin{equation} 1 \le (1+\epsilon_{n_k} x/n_k)^{n_k} \le e^{\epsilon_{n_k} x} \end{equation}

Then take the limit $\epsilon_{n_k} \rightarrow 0$. For the $\epsilon_{n_k} \le 0$ subsequence, since $f(y) = (1+y)^n, |y| < 1$ is convex:

\begin{equation} 1 \ge (1 + \epsilon_{n_k} x/n_k)^{n_k} \ge 2 - (1 - \epsilon_{n_k}x/n_k)^{n_k}, n_k \text{ large enough} \end{equation}

where we used using $(1/2)f(y) + (1/2)f(-y) \ge f(0)$. We get the same limit, therefore the whole sequence converges to 1.

Jim
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Yes, your solution is correct.

Alternate Method -

Let $y=\lim\limits_{n\rightarrow\infty}\left(1+\epsilon_n \frac{x}{n}\right)^n$. This means, $$\ln(y)=\lim\limits_{n\rightarrow\infty}\left(\ln(1+\epsilon_n \frac{x}{n})\cdot\frac{1}{n^{-1}}\right)\approx\lim\limits_{n\rightarrow\infty}\left(\left(\epsilon_n \frac{x}{n}+O\left((e_n\frac{x}{n})^2\right)\right)\cdot\frac{1}{n^{-1}}\right)=\lim\limits_{n\rightarrow\infty}(x\epsilon_n+nO\left((e_n\frac{x}{n})^2\right))=0$$ Hence, $y=1$.

  • No, the OPs solution is not correct. And this is essentially correct, but you need to be more precise about what $\approx$ means. – Thomas Andrews Jun 18 '21 at 07:23
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$$ \lim_{n\to +\infty}\left(1+\varepsilon_n \frac xn\right)^n=\lim_{n\to +\infty}\left(1+\frac{x}{n/\varepsilon_n}\right)^n=\lim_{n\to +\infty}\left[\left(1+\frac{x}{n/\varepsilon_n}\right)^{n/\varepsilon_n}\right]^{\varepsilon_n}=\lim_{n\to\infty}(e^x)^{\varepsilon_n}=1. $$

Note: This strategy works if $\varepsilon_n$ does not change sign, for $n > n_0$.

PierreCarre
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  • To claim $\lim_{n \to +\infty} (1 + \frac {x}{n/ \epsilon_n})^{n/ \epsilon_n}$ is equal to $e^x$ don't you first need to show $n/ \epsilon_n \to \infty$ as $n \to \infty$? – David Choi Jun 18 '21 at 09:07
  • @DavidChoi Yes, you are correct... This approach works only if $\varepsilon_n >0$ (for $n>n_0$) because we need to be sure that $n/\varepsilon_n \to +\infty$. – PierreCarre Jun 18 '21 at 09:12