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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)^uright)^r \
&= left(lim_{u rightarrow infty} left(1 + frac{1}{u}right)^uright)^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)^uright)^{-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.

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

Consider proving the reciprocal, i.e.
$$
lim left(1 - frac 1nright)^{-n} = mathrm e.
$$
The expression inside could be rewritten as
$$
left(1 - frac 1n right)^{-n} = left(frac {n-1}nright)^{-n} = left(frac n{n-1}right)^n = left(1 + frac 1{n-1}right)^n = left(1 + frac 1{n-1}right)^{n-1} cdot left(1 + frac 1{n-1}right).
$$
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 $$begin{align*} limlimits_{n rightarrow infty} ln P & = limlimits_{n rightarrow infty} frac {frac {1} {left (1 - frac 1 n right )} cdot frac {1} {n^2}} {-frac {1} {n^2}}. \ & = - limlimits_{n rightarrow infty} frac {1} {left (1 - frac 1 n right )}. \ & = -1. end{align*}$$

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

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

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

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

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

Then

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

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

$exp (-1)$.

Used: Continuity of $exp$.