How do you find the limit of (1 - 1/x)^x(11x)x as x approaches infinity?

3 Answers
Jul 4, 2015

The limit is 1/e1e

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

Aug 20, 2017

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).

Aug 20, 2017

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