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