# # lim_(n->oo)(1+1/n)^n = # ?

##### 3 Answers

Let

It is obvious that:

We'll show that sequence

Using Bernoulli's inequality

So, the sequence

Using Bernoulli's inequality again:

it follows that

Using inequality of arithmetic and geometric means:

it follows:

and

It follows that

Finally,

Let some value

#lnx = ln(lim_(n->oo) (1+1/n)^n)#

#lnx = lim_(n->oo)(nln(1+1/n))#

#lnx = lim_(n->oo)(ln(1+1/n)/(1/n))#

By inspection, we've converted the right side into a

#lnx = lim_(n->oo)((1/(1+1/n)*cancel(-1/n^2))/cancel(-1/n^2))#

#lnx = lim_(n->oo)(1/(1+1/n))#

wherein the

#lnx = lim_(nrarroo)(1/(1+0))#

#lnx=lim_(nrarroo)1#

#lnx=1#

Thus, we undo the natural logarithm to get:

#color(blue)(lim_(n->oo) (1+1/n)^n)#

#= e^(ln(lim_(n->oo) (1+1/n)^n))#

# = e^(lnx) = x = e^1 = color(blue)(e)# .

so

Finally

and