**Definition used:**

A sequence
{an}
has a limit *L*, if every
ε>0
there exists an integer *N* such that
|an−L|<ε for n>N
. That is,
limn→∞an=L
.

**Proof:**

Given that,
limn→∞ an=0
and
{bn}
is bounded.

Then it is enough to prove that, for given
ε>0
and there is some integer *N* such that
|anbn|<ε if n>N
.

Since
{bn}
is bounded, there is some integer *M* such that
|bn|≤M
.

Multiply by
|an|
on both the sides of the inequality.

|an||bn|≤|an|M
(1)

Since
limn→∞ an=0
and let
εM>0
, by the definition stated above,

|an−0|<εM for all n>N

That is,
|an|<εM for all n>N
(2)

For all n>N
,

|anbn−0|=|anbn|=|an||bn|≤|an|⋅M [By equation (1)]<εM⋅M [By equation (2)]

=ε

Therefore,
|an−0|<ε for all n>N
.

Since
ε
is any arbitrary element,
limn→∞(anbn)=0
.

Hence the required proof is obtained.