Question

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

Answer 1

The limit is `1/e`

Explanation:

`lim_(xrarroo)(1-1/x)^x` has the form `1^oo` which is an indeterminate form.

We will use logarithms and the exponential function.

Now,
`(1-1/x)^x = e^(ln(1-1/x)^x)`

So we will investigate the limit of the exponent.

`lim_(xrarroo)(ln(1-1/x)^x)`

It will be convenient to note that: `1-1/x = (x-1)/x`

`ln(1-1/x)^x = ln ((x-1)/x)^x = xln((x-1)/x)`

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

Now as `xrarroo` we get the form `oo * ln1 = oo*0` So we'll put the reciprocal of one of these in the denominator so we can use l'Hopital's Rule.

`xln((x-1)/x) = (ln((x-1)/x))/(1/x)` Now, as `xrarroo` we get the form `0/0` Apply l'Hopitals's Rule:

The more tedious derivative is:
`d/dx(ln((x-1)/x)) = 1/((x-1)/x)*d/dx((x-1)/x)`

`= x/(x-1) * 1/x^2 = 1/(x(x-1))`

So we get from `(ln((x-1)/x))/(1/x)`

to `(1/(x(x-1)))/(-1/x^2) = - x/(x-1)`

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

`lim_(xrarroo)(- x/(x-1)) = -1`

Summary:

`(1-1/x)^x = e^(ln(1-1/x)^x)`

And as `xrarroo` the exponent goes to `-1`

Therefore:

`lim_(xrarroo)(1-1/x)^x = lim_(xrarroo)e^(ln(1-1/x)^x) = e^-1 = 1/e`

Answer 2

`lim_(x -> \infty) (1-1/x)^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 -> \infty) (1+a/x)^x = e^a`.

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

`lim_(x -> \infty) (1-1/x)^x = e^(-1)`.

Answer 3

This solution assumes that we know that `lim_(urarr0)(1+u)^(1/u) = e`

Explanation:

`(1-1/x)^x = (1+(-1/x))^x = ((1+(-1/x))^-x)^-1`

Let `u = -1/x` and note that as `xrarroo`, `urarr0^-`

so

`lim_(xrarroo)(1-1/x)^x = lim_(urarr0^-)((1+u)^(1/u))^-1`

`= (lim_(urarr0)(1+u)^(1/u))^-1`

`= e^-1 = 1/e`