Given:
The sequence is
an=nsin(1n)
.
Definition used:
If
an
is a sequence and
limn→∞an
exists, then the sequence
an
is said to be converges; otherwise it diverges.
Limit chain rule:
“If
limu→bf(u)=L
and
limx→ag(x)=b
with
f(x)
is said to be 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 converges or diverges.
Compute the value of
limn→∞an=limn→∞(nsin(1n))
.
limn→∞an=limn→∞(nsin(1n)) =limn→∞(sin(1n)1n)=sin(1∞)1∞=sin(0)0
Since
00
is in indeterminate form, apply L’Hospital’s rule.
limn→∞(nsin(1n))=limn→∞(ddn(sin(1n))ddn(1n)) =limn→∞(ddn(sin(1n))⋅ddn(1n)ddn(1n)) =limn→∞(cos(1n)⋅−1n2−1n2) =limn→∞(cos(1n))
Consider the functions
g(n)=1n
and
f(u)=cos(u)
and obtain
limn→∞g(n)
.
limn→∞g(n)=limn→∞(1n) =1∞ =0
Obtain
limu→0f(u)
.
limu→0f(u)=limu→0(cos(u)) =cos(0) =1
Apply the limit chain rule and obtain the limit.
Since
g(n)=1n
and
f(u)=cos(u)
,
limn→∞f(1n)=1
and
limn→∞(cos(1n))=1
.
Therefore,
limn→∞(cos(1n))=1
.
Since
limn→∞an
exists and by using the definition of the sequence, it can be concluded that the sequence converges to the limit 1.
Therefore, the sequence converges to the limit 1.