0

Define $$e=\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n,$$ how to prove that $$e^x=\lim_{n\to\infty}\left(1+\dfrac{x}{n}\right)^n,$$ for all $x$. Thank you.

kong
  • 75
  • 2

3 Answers3

1

Use the definition

$$ e = \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n $$

Then

$$ e^x = \left( \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n \right)^x = \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^{nx} = \lim_{n \rightarrow \infty} \left( 1 + \frac{x}{nx} \right)^{nx} = \lim_{n \rightarrow \infty} \left( 1 + \frac{x}{n} \right)^{n} $$

johannesvalks
  • 6,587
  • Can you justify each equality by mentioning the theorems which you used? – Chellapillai Jul 05 '14 at 10:23
  • That last step where you draw the power of x into the brackets to attain $(1+x/n)$, how is that true? – Just_a_fool Jul 05 '14 at 10:23
  • Write $\frac{1}{n}$ as $\frac{x}{xn}$ and then use that $nx$ goes to infinity as well - so replace $nx$ by $n$... – johannesvalks Jul 05 '14 at 10:26
  • @Chellapillai: I have no idea how those theorems are called - we learn them we use them and we forget the names... – johannesvalks Jul 05 '14 at 10:28
  • Okay. Just I would like to know that whether we have to use the fact that $y\mapsto y^x(y\in[0,\infty))$ is continuous or something else? – Chellapillai Jul 05 '14 at 10:41
  • Replacing $nx$ by $n$ is not allowed here as $n$ is an integer and $nx$ may not be an integer. – Paramanand Singh Jul 05 '14 at 12:01
  • It is nowhere written that $n$ is an integer. It is a limit that goes to infinity, and both $n$ and $nx$ go to infinity (for $x \ne 0$... – johannesvalks Jul 05 '14 at 12:07
  • @johannesvalks: By convention, $n\to\infty$ implies that $n$ is a positive integer unless explicitly stated otherwise. – Paramanand Singh Jul 05 '14 at 16:13
  • @Paramanand: Infinity is not a number, it is a concept. Therefore infinity is never an integer. – johannesvalks Jul 05 '14 at 16:18
  • @johannesvalks: I am not saying that $\infty$ is an integer. I am saying that $n$ is an integer. – Paramanand Singh Jul 05 '14 at 16:27
  • @Paramanand: $\lim_{x \rightarrow\infty} x = \lim_{x \rightarrow\infty} \lfloor x \rfloor \Big( 1 + \frac{x-\lfloor x \rfloor}{\lfloor x \rfloor} \Big) = \lim_{x \rightarrow\infty} \lfloor x \rfloor = \lim_{\lfloor x \rfloor \rightarrow \infty} x$ – johannesvalks Jul 05 '14 at 16:29
  • @Paramanand: it is irrelevant for the limit that goes to infinity to distinguish integers from non integers in this case. – johannesvalks Jul 05 '14 at 16:35
  • @johannesvalks: you seem to suggest that when $n\to\infty$ then it does not matter that $n$ is an integer or not. But it does matter! $\lim_{n\to\infty}\sin n\pi = 0$ if $n$ is positive integer. If $n$ is not a positive integer then this limit does not exist. – Paramanand Singh Jul 05 '14 at 16:37
  • @Paramanand: You have missed my words "in this case". So examples of other cases are irrelevant. – johannesvalks Jul 05 '14 at 16:39
  • 1
    @johannesvalks: I know that in the current question it does not matter if $n$ is integer or real, but this fact is not obvious and very difficult to prove. Hence I want to highlight the issue. – Paramanand Singh Jul 05 '14 at 16:39
  • @Paramanand: Now we are talking! This is a good reply. In this case it is clear, and you ask to 'prove' it... – johannesvalks Jul 05 '14 at 16:42
  • Let us continue this discussion in chat. – Paramanand Singh Jul 05 '14 at 16:43
  • @johannesvalks: you should see my post http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-2_10.html in order to understand why it is not clear/obvious and why it needs a proof. – Paramanand Singh Jul 05 '14 at 16:50
  • @Paramanand: Now for me the chat is new... I went there but how do you reply? – johannesvalks Jul 05 '14 at 16:50
  • @Paramanand: What you wrote on you blog is based on $n$ as integers. – johannesvalks Jul 05 '14 at 17:25
  • @johannesvalks: yes in that post $n$ is an integer and towards the end of http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-3.html i show the case when $n$ is real. Here I use variable $t$ instead of $n$. (see equation $(14)$ of this post). – Paramanand Singh Jul 05 '14 at 17:37
1

Let $\frac{n}{x}=t$. Then as $n$ goes to infinity, $t$ does too. Therefore by considering the given definition of $e$ we have

$$\lim_{n\to\infty}\left(1+\dfrac{x}{n}\right)^n=\lim_{t\to\infty}\left(1+\dfrac{1}{t}\right)^{xt}=(\lim_{t\to\infty}\left(1+\dfrac{1}{t}\right)^{t})^x=e^x$$

Fermat
  • 5,366
-1

By applying $\log$ and using L'Hospital's rule $$\lim_{n \to \infty}\frac{\log\left(1+\frac{x}{n}\right)}{\frac{1}{n}}=\lim_{n\to \infty}\frac{\frac{1}{1+\frac{x}{n}}\left(\frac{-x}{n^2}\right)}{\frac{-1}{n^2}}=x$$

tattwamasi amrutam
  • 13,040