**Given:**

The sequence is
a1=2, an+1=13−an
(1)

**Theorem used: Monotonic Sequence Theorem**

“If the sequence is bounded and monotonic, then the sequence is convergent.”

**Proof: Induction Method**

Claim: Let
Pn
be the statement that
an+1≤an
and
0<an≤2
. (2)

*Base case:*
n=1

Substitute 1 for *n* in equation (2),

a1+1≤a1 and 0<a1≤2a2≤a1 and 0<a1≤2

Obtain the value of
a2
.

Substitute 1 for *n* in equation (1),

a1+1=13−a1a2=13−2 [∵a1=2]=11=1

Thus, the value of
a2=1
.

Since the value of
a1=2 and a2=1
,
a2≤a1 and 0<a1≤2
.

Therefore, the claim is true for
n=1
.

*Induction hypothesis:*
n=k

Assume that the claim is true when
n=k
.

If
Pk
is the statement, then
ak+1≤ak
and
0<ak≤2
is true.

*Inductive step:*
n=k+1

To prove that the claim is true when
n=k+1

That is, if
Pk+1
is the statement, then
a(k+1)+1≥ak+1
and
0<ak+1≤2
is true.

Substitute
k+1
for *n* in equation (1),

a(k+1)+1=13−ak+1
(3)

From induction hypothesis,

ak+1≤ak−ak+1>−ak

Add by 3 on both sides of the inequality.

3−ak+1>3−ak

13−ak+1≤13−ak
(4)

Substitute the inequality (4) in the equation (3) and then plug in
13−ak
for
ak+1
.

a(k+1)+1≤13−ak

a(k+1)+1≤ak+1
(5)

Consider
ak+1
, then apply equation (1) and the induction hypothesis
0<ak≤2
.

ak+1=13−ak≤13−2=1<2

Therefore,
0<ak+1≤2
. (6)

From (5) and (6),
a(k+1)+1≤ak+1 and 0<ak+1≤2
.

Hence, the claim is true for all *n* and
0<an≤2 for n≥1
.

Since the sequence
{an}
is bounded and monotonic and by Monotonic Sequence Theorem, the sequence is convergent.

That is,
limn→∞an
exists.

Obtain the limit of the sequence.

limn→∞an+1=limn→∞(13−an)=1limn→∞(3)−limn→∞(an)=13−limn→∞an

Since the sequence
{an}
is convergent,
limn→∞an=L
.

Thus,
limn→∞an+1=L
.

Substitute *L* for
limn→∞an and limn→∞an+1
.

L=13−L1=L(3−L)1=3L−L2

Rearrange the terms of the equation and form the quadratic equation such that
L2−3L+1=0
.

Solve the equation for *L* by using the q Quadratic formula.

L=−(−3)±(−3)2−4⋅1⋅12⋅1=3±9−42=3±52

Since
0<an≤2 for all n≥1
, the value of
L=3−52
.

Therefore, the limit of the sequence is
3−52
.