Limit for $e$ and $\frac{1}{e}$

My question concerns the derivation of this: $$e^r = \lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n \ \ ...(1).$$ One of the definitions of $$e$$ is as follows: $$e = \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n.$$

Then, textbooks usually derive equation (1) in the following manner:

\begin{align} \lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^{ru} \ \ \text{where} \ u = \frac{n}{r}\\ &= \lim_{u \rightarrow \infty} \left(\left(1 + \frac{1}{u}\right)^u\right)^r \\ &= \left(\lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^u\right)^r \\ &= e^r. \end{align}

This argument is fine if $$r > 0$$ since $$u \rightarrow \infty$$ as $$n \rightarrow \infty$$, but when $$r < 0$$, $$u \rightarrow - \infty$$ as $$n \rightarrow \infty$$.

How can I extend the proof for (1) where $$r$$ is any real number?

When $$r = 0$$, $$\lim_{n \rightarrow \infty} (1+0/n)^n = 1$$ (Although, I should be careful about evaluating limits that look like $$1^{\infty}"$$.)

Here's my attempt so far for the case where $$r < 0$$: \begin{align} \lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^{-ru} \ \text{where} \ u = -\frac{n}{r} \\ &= \left(\lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^u\right)^{-r}. \end{align}

My question boils down to how to show the following limit from the definition above for $$e$$ $$\lim_{n \rightarrow \infty} \left(1 - \frac{1}{n}\right)^n = \frac{1}{e}.$$ Thanks.

• Isn't the last limit $\lim\limits_{n \rightarrow \infty} \left (1 - \frac 1 n \right )^n = \frac {1} {e}$? Commented Apr 14, 2019 at 16:31
• The substitution is still valid for $r\lt0$ because the bracket contains a number $\lt1$. Commented Apr 14, 2019 at 16:34
• Side note: no need to be careful for $r=0$. You have $(1+0/n)^n=1$ for all $n$.
– Dirk
Commented Apr 14, 2019 at 19:51

\begin{align*}\lim_{n\to\infty}\left( 1 - \frac{1}{n} \right)^n &= \lim_{n\to\infty}\left(\frac{n-1}{n}\right)^n = \lim_{n\to\infty}\left(\frac{1}{\frac{n}{n-1}}\right)^n = \lim_{n\to\infty}\frac{1}{\left(\frac{n}{n-1}\right)^n} \\ &= \lim_{n\to\infty}\frac{1}{\left(1 + \frac{1}{n-1}\right)^n} = \lim_{n\to\infty} \left(\frac{1}{\left(1 + \frac{1}{n-1}\right)^{n-1}} \cdot \frac{1}{1 + \frac{1}{n-1}} \right) \\ &= \lim_{n\to\infty} \frac{1}{\left(1 + \frac{1}{n-1}\right)^{n-1}} = \lim_{n\to\infty} \frac{1}{\left(1 + \frac{1}{n}\right)^{n}} = \frac{1}{e} \\ &= e^{-1} \end{align*}

Consider proving the reciprocal, i.e. $$\lim \left(1 - \frac 1n\right)^{-n} = \mathrm e.$$ The expression inside could be rewritten as $$\left(1 - \frac 1n \right)^{-n} = \left(\frac {n-1}n\right)^{-n} = \left(\frac n{n-1}\right)^n = \left(1 + \frac 1{n-1}\right)^n = \left(1 + \frac 1{n-1}\right)^{n-1} \cdot \left(1 + \frac 1{n-1}\right).$$ Now use the definition of $$\mathrm e$$: $$\lim_n \left(1 + \frac 1{n-1}\right)^{n-1} = \mathrm e.$$ Since $$\lim_n \left(1 + \frac 1{n-1}\right) = 1,$$ we conclude that $$\lim_n \left(1 - \frac 1n \right)^{-n} = \lim_n \left(1 + \frac 1{n-1}\right)^{n-1} \cdot \lim_n \left(1 + \frac 1{n-1}\right) = \mathrm e \cdot 1 = \mathrm e.$$ by the law of arithmetic operations of limits.

Therefore the original limit is $$1/\mathrm e$$.

Let $$P= \left (1 - \frac 1 n \right )^n.$$ Then $$\ln P = \frac {\ln \left (1 - \frac 1 n \right )} {\frac 1 n}.\ (*)$$

Now as $$n \rightarrow \infty$$ then $$(*)$$ is a $$\frac 0 0$$ form. Applying L'Hospital we have \begin{align*} \lim\limits_{n \rightarrow \infty} \ln P & = \lim\limits_{n \rightarrow \infty} \frac {\frac {1} {\left (1 - \frac 1 n \right )} \cdot \frac {1} {n^2}} {-\frac {1} {n^2}}. \\ & = - \lim\limits_{n \rightarrow \infty} \frac {1} {\left (1 - \frac 1 n \right )}. \\ & = -1. \end{align*}

This shows that $$\lim\limits_{n \rightarrow \infty} P =\lim\limits_{n \rightarrow \infty} \left (1 - \frac 1 n \right )^n = \frac 1 e.$$

The easiest way, working for all $$r \in \mathbb{R}$$, is to write $$\left( 1+ \frac{r}{n}\right)^n =\exp \left( n \ln \left( 1+ \frac{r}{n}\right)\right) = \exp \left( n \left( \frac{r}{n} + o\left( \frac{1}{n}\right) \right)\right) = \exp \left( r + o\left( 1\right) \right)$$

So it converges to $$\exp(r)$$.

$$f(x):=\log x;$$

$$f'(1)= \lim_{n \rightarrow \infty}\dfrac{f(1-1/n)-f(1)}{-(1/n)}$$;

Hence

$$\lim_{n \rightarrow \infty} (n\log(1-1/n)-\log 1)=$$

$$\lim_{n \rightarrow \infty} (-\dfrac{\log (1-1/n)-\log 1}{-(1/n)})=- \log '(1)=-1$$

Then

$$\lim_{n \rightarrow \infty} \exp n(\log (1-1/n))=$$

$$\exp (\lim_{n \rightarrow \infty} n\log (1-1/n))=$$

$$\exp (-1)$$.

Used: Continuity of $$\exp$$.