Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

One of the standard definitions of $e$ is as

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

But in all cases I've seen this limit, it is proven as a limit of the sequence $\Big\{\big(1 + \frac{1}{n}\big)^n\Big\}$, which seems to cover the limit for only $n$ as an integer. Now my question is whether this sequence limit is equivalent to a normal limit for which $n$ can be any real number. I can think of cases which this isn't generally true;

$$\lim_{n\rightarrow\infty}\ \sin(n\pi)$$

comes readily to mind, for which the limit as a sequence is simply $0$ but as a general limit, it is undefined. The limit for $e$ is used exactly as if it were a normal limit, which leads me to believe it is equivalent. Are there conditions for which the limit of a sequence and the corresponding function are identical?

share|improve this question
3  
+1. This is an insightful question, IMO. –  JavaMan Oct 20 '11 at 23:54
    
If $\lim_x f(x)$ exists, then $\lim_x f(x) = \lim_n f(n)$ (but as you point, there are cases where $\lim_n f(n)$ exists but $\lim_x f(x)$ does not). –  Joel Cohen May 13 '12 at 19:32
add comment

1 Answer

up vote 13 down vote accepted

If $x>0$ be a real number, then write $x=n+y_x$ where $0 \leq y_x \leq 1$.

It is easy then to show that

$$ \left(1+\frac{1}{n+1} \right)^n \leq \left(1+\frac{1}{x}\right)^x \leq \left(1+\frac{1}{n}\right)^{n+1} \,.$$

Using this, you can prove that if

$$e=\lim_n \left(1+\frac{1}{n}\right)^n$$ then $$\lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x =e$$

Edit

To complete the answer, in general if $f(x)$ is monotonic, then for sure

$$\lim_n f(n)= \lim_x f(x) \,.$$

I don't recall if $\left(1+\frac{1}{x}\right)^x$ is monotonic (and I am too lazy to differentiate it), but what we showed above is that it is at least "close to being monotonic". What I mean by this, I showed that $$f(n) h(n) \leq f(x) \leq f(n+1)g(n+1) \forall x\in [n, n+1)$$ where $h(n)$ and $g(n)$ go to $1$. The you basically sqeeze it.

share|improve this answer
    
$(1+\frac{1}{x})^x$ increases monotonicly for positive $x$ –  Henry Oct 20 '11 at 23:28
    
@DJC: Good idea, but since both $\log\left(1+\frac{1}{x}\right)$ and $\frac{1}{x+1}$ are less than $\frac{1}{x}$, I think you need to go a bit farther in the asymptotic expansion: $\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2x^2}+O\left(\frac{1}{x^3}\‌​right)$ and $\frac{1}{x+1}=\frac{1}{x}-\frac{1}{x^2}+O\left(\frac{1}{x^3}\right)$. And even then, we need to consider when $O\left(\frac{1}{x^3}\right)$ is small enough. –  robjohn Oct 21 '11 at 0:16
2  
Since $e^u\ge1+u$ for all $u\in\mathbb{R}$, we also have $u\ge\log(1+u)$ for $u>-1$. The derivative of the logarithm of $(1+1/x)^x$ is $$ \begin{align} \log\left(1+\frac{1}{x}\right)-\frac{1}{x+1} &=-\log\left(1-\frac{1}{x+1}\right)-\frac{1}{x+1}\\ &=u-\log(1+u)\\ &\ge0 \end{align} $$ where $u=-\frac{1}{x+1}>-1$ for $x>0$. Therefore, $(1+1/x)^x$ is monotonically increasing for $x>0$. –  robjohn Oct 21 '11 at 0:41
    
Your sandwich between $\left(1+\frac{1}{n+1} \right)^n$ and $\left(1+\frac{1}{n}\right)^{n+1}$ is similar to this sandwich. (+1) –  robjohn Oct 21 '11 at 2:13
    
@robjohn: You're absolutely right. Thanks for cleaning that up. –  JavaMan Oct 21 '11 at 4:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.