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

Answer 1
Let #a_n# and #b_n#, #n\inN#, are defined as:
#a_n=(1+1/n)^n#,
#b_n=(1+1/n)^(n+1) = a_n(1+1/n).#

It is obvious that:

#lim_(n->infty) a_n = (lim_(n->infty) b_n)/(lim_(n->infty) (1+1/n)) = lim_(n->infty) b_n.#
We'll show that sequence #b_n# is decreasing.
#b_(n-1)/b_n = (1+1/(n-1))^n/(1+1/n)^(n+1) = (n/(n-1))^n/((n+1)/n)^(n+1)=n^(2n+1)/((n-1)^n(n+1)^(n+1))#
#b_(n-1)/b_n = (n^2/(n^2-1))^n*n/(n+1) = (1+1/(n^2-1))^n*n/(n+1)#
Using Bernoulli's inequality #(1+x)^n > 1+nx# it follows:
#(1+1/(n^2-1))^n > 1+n/(n^2-1)# and:
#b_(n-1)/b_n > (1+n/(n^2-1))*n/(n+1) > (1+n/n^2)*n/(n+1)#
#b_(n-1)/b_n > (n(n+1))/n^2*n/(n+1) = 1#
So, the sequence #b_n# is decreasing.

Using Bernoulli's inequality again:

#b_n = (1+1/n)^(n+1) > 1+(n+1)/n = 1+1+1/n = 2+1/n > 2#
it follows that #b_n# has a lower bound. #b_n# is convergent.

Using inequality of arithmetic and geometric means:

#root(n+1)((1+1/n)^n*1) < (n(1+1/n)+1)/(n+1) = (n+2)/(n+1)#

it follows:

#(1+1/n)^n < ((n+2)/(n+1))^(n+1) = (1+1/(n+1))^(n+1),#
#a_n < a_(n+1)#
and #a_n# is increasing.
#a_n = (1+1/n)^n < (1+1/n)^(n+1) = b_n#
It follows that #a_n# has an upper bound (because #b_n# is decreasing and has a lower bound). Sequence #a_n# is increasing and has an upper bound so it's convergent.

Finally,

#e = lim_(n->infty) (1+1/n)^n.#
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Answer 2
Let some value #x = lim_(n->oo) (1 + 1/n)^n#, where #n# is an arbitrary variable (i.e. does not have to be an integer!). Then the fastest way to proceed is as follows.
#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 #0/0# form. Therefore, we can use L'Hopital's rule.
#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 #1/n# term vanishes as #n->oo#.
#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)#.
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Answer 3
#(1+1/n)^n < (1+1/(n+1))^(n+1)# is increasing for #n->oo#
#(1+1/n)^n = 1+1/(1!)n/n+1/(2!)(n/n)((n-1)/n) + 1/(3!)(n/n)((n-1)/n)((n-2)/n)+cdots#
#((n-1)/n)(1+1/n)^n le 1-1+ (n-1)/n+1/(1!)(n-1)/n+1/(2!)((n-1)/n)^2+1/(3!)((n-1)/n)^3 + 1/(4!)((n-1)/n)^4+cdots +#
#((n-1)/n)(1+1/n)^n le (n-1)/n-1+sum_(k=0)^n1/(k!)((n-1)/n)^k#

so

#lim_(n->oo)((n-1)/n)(1+1/n)^n le lim_(n->oo)(-1/n)+lim_(n->oo)sum_(k=0)^n1/(k!)((n-1)/n)^k lelim_(n->oo)( -1/n)+lim_(n->oo)sum_(k=0)^n1/(k!) #

Finally

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

and

# lim_(n->oo)(1+1/n)^n le e#
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Answer 4

The limit of ( (1 + \frac{1}{n})^n ) as ( n ) approaches infinity is equal to ( e ), where ( e ) is Euler's number, approximately ( 2.71828 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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