# How do you find the limit of (1 - 1/x)^x as x approaches infinity?

Jul 4, 2015

The limit is $\frac{1}{e}$

#### Explanation:

${\lim}_{x \rightarrow \infty} {\left(1 - \frac{1}{x}\right)}^{x}$ has the form ${1}^{\infty}$ which is an indeterminate form.

We will use logarithms and the exponential function.

Now,
${\left(1 - \frac{1}{x}\right)}^{x} = {e}^{\ln {\left(1 - \frac{1}{x}\right)}^{x}}$

So we will investigate the limit of the exponent.

${\lim}_{x \rightarrow \infty} \left(\ln {\left(1 - \frac{1}{x}\right)}^{x}\right)$

It will be convenient to note that: $1 - \frac{1}{x} = \frac{x - 1}{x}$

$\ln {\left(1 - \frac{1}{x}\right)}^{x} = \ln {\left(\frac{x - 1}{x}\right)}^{x} = x \ln \left(\frac{x - 1}{x}\right)$

(Using a property of logarithms to bring the exponent down)

Now as $x \rightarrow \infty$ we get the form $\infty \cdot \ln 1 = \infty \cdot 0$ So we'll put the reciprocal of one of these in the denominator so we can use l'Hopital's Rule.

$x \ln \left(\frac{x - 1}{x}\right) = \frac{\ln \left(\frac{x - 1}{x}\right)}{\frac{1}{x}}$ Now, as $x \rightarrow \infty$ we get the form $\frac{0}{0}$ Apply l'Hopitals's Rule:

The more tedious derivative is:
$\frac{d}{\mathrm{dx}} \left(\ln \left(\frac{x - 1}{x}\right)\right) = \frac{1}{\frac{x - 1}{x}} \cdot \frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x}\right)$

$= \frac{x}{x - 1} \cdot \frac{1}{x} ^ 2 = \frac{1}{x \left(x - 1\right)}$

So we get from $\frac{\ln \left(\frac{x - 1}{x}\right)}{\frac{1}{x}}$

to $\frac{\frac{1}{x \left(x - 1\right)}}{- \frac{1}{x} ^ 2} = - \frac{x}{x - 1}$

Now we can probably find this limit without l'Hopital:

${\lim}_{x \rightarrow \infty} \left(- \frac{x}{x - 1}\right) = - 1$

Summary:

${\left(1 - \frac{1}{x}\right)}^{x} = {e}^{\ln {\left(1 - \frac{1}{x}\right)}^{x}}$

And as $x \rightarrow \infty$ the exponent goes to $- 1$

Therefore:

${\lim}_{x \rightarrow \infty} {\left(1 - \frac{1}{x}\right)}^{x} = {\lim}_{x \rightarrow \infty} {e}^{\ln {\left(1 - \frac{1}{x}\right)}^{x}} = {e}^{-} 1 = \frac{1}{e}$

Aug 20, 2017

${\lim}_{x \to \setminus \infty} {\left(1 - \frac{1}{x}\right)}^{x} = {e}^{- 1}$.

#### Explanation:

Note that the exponential function has the power series e^a = sum_(n=0)^(\infty) a^n/(n!).

(1+a/x)^x=sum_(n=0)^(x) (x!)/(n!(x-n)!)(a/x)^n,
lim_(x -> \infty) (1+a/x)^x = lim_(x -> \infty) sum_(n=0)^(x) ((x!)/(x^n(x-n)!)) a^n/(n!) .

Consider,

(x!)/(x^n (x-n)!) = ((x)(x-1)...(x-n+1))/(x^n),
(x!)/(x^n (x-n)!) = (x/x) ((x-1)/(x))... ((x-n+1)/(x))

Or more formally,

(x!)/(x^n(x-n)!) = prod_(k=1)^(n-1) (x-k)/(x)

Then,

lim_(x \to \infty) (x!)/(x^n(x-n)!) = prod_(k=1)^(n-1) lim_(x \to \infty) (x-k)/(x),
lim_(x \to \infty) (x!)/(x^n(x-n)!) = prod_(k=1)^(n-1) ( 1 - lim_(x \to \infty) k/(x)),
lim_(x \to \infty) (x!)/(x^n(x-n)!) = prod_(k=1)^(n-1) 1,
lim_(x \to \infty) (x!)/(x^n(x-n)!) = 1.

Then,

lim_(x -> \infty) (1+a/x)^x = sum_(n=0)^(\infty) a^n/(n!) ,
${\lim}_{x \to \setminus \infty} {\left(1 + \frac{a}{x}\right)}^{x} = {e}^{a}$.

We conclude that, with $a = - 1$,

${\lim}_{x \to \setminus \infty} {\left(1 - \frac{1}{x}\right)}^{x} = {e}^{- 1}$.

Aug 20, 2017

This solution assumes that we know that ${\lim}_{u \rightarrow 0} {\left(1 + u\right)}^{\frac{1}{u}} = e$

#### Explanation:

${\left(1 - \frac{1}{x}\right)}^{x} = {\left(1 + \left(- \frac{1}{x}\right)\right)}^{x} = {\left({\left(1 + \left(- \frac{1}{x}\right)\right)}^{-} x\right)}^{-} 1$

Let $u = - \frac{1}{x}$ and note that as $x \rightarrow \infty$, $u \rightarrow {0}^{-}$

so

${\lim}_{x \rightarrow \infty} {\left(1 - \frac{1}{x}\right)}^{x} = {\lim}_{u \rightarrow {0}^{-}} {\left({\left(1 + u\right)}^{\frac{1}{u}}\right)}^{-} 1$

$= {\left({\lim}_{u \rightarrow 0} {\left(1 + u\right)}^{\frac{1}{u}}\right)}^{-} 1$

$= {e}^{-} 1 = \frac{1}{e}$