**Solution:**

The sequence is converges and its limit is 1.

**Given:**

The sequence is
an=e−1n
.

**Definition used:**

If
an
is sequence and
limn→∞an
is exists then the sequence
an
is convergent otherwise it is divergent

**Result used: Limit chain rule**

If
limu→bf(u)=L
and
limx→ag(x)=b
with
f(x)
is continuous at
x=b
, then the value of
limx→af(g(x))
is *L*.

**Calculation:**

Obtain the limit of the sequence to investigate whether the sequence is converges or diverges.

Compute the value of
limn→∞an=limn→∞e−1n
.

Let
g(n)=−1n
and
f(u)=eu
.

Obtain
limn→∞g(n)
.

limn→∞g(n)=limn→∞(−1n) =−limn→∞(1n) =−1∞ =0

Obtain
limu→0f(u)
.

limu→0f(u)=limu→0(eu) =e0 =1

Apply limit chain rule and obtain the limit.

Since
f(u)=eu
and
g(n)=−1n
,
limn→∞eg(n)=1
and
limn→∞e−1n =1
.

Therefore,
limn→∞f(g(n))=1
.

Since
limn→∞an
is exist and by using the definition of the sequence, it can be concluded that the sequence is converges to the limit 1.

Since the sequence
an
exists as *n* tends to infinity, and by using the definition the sequence is convergent with limit is 1.

Therefore, the sequence is converges to the limit 1.