What is the limit of #(1+1/x)^x# as x approaches infinity?

Question

5830 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-07T18:00:04

#e#

We are trying to find:

#lim_(x->oo)[(1+1/x)^x]#

Let's begin by expanding the term in brackets using the binomial theorem

#(1+1/x)^x=1+((x),(1))1/x+((x),(2))1/x^2 +...+((x), (x))1/x^x #

such that the #k^(th)# term in the series is

#((x), (k))1/x^k=(x!)/(k!(x-k)!)*1/x^k#

We can simplify this expression by cancelling terms with #x# in them and gathering them together. Then we take the limit of each term in the sum

#lim_(x->oo)[1/(k!)*(x(x-1)(x-2)...(x-k+1))/(x^k)]#

From this we can see that the term on the right goes to 1 as #x->oo#. So the sum becomes:

#1/(0!)+1/(1!)+1/(2!) + ... #

This is the known series for the natural number, #e#, therefore

#lim_(x->oo)[(1+1/x)^x] = e#

Interestingly, this limit sometimes used to define #e#. For more information, see the "Series for e" section at the following link:
https://en.wikipedia.org/wiki/Binomial_theorem