Whether the sequence converges or diverges and obtain the limit if the sequence converges.

The sequence converges to the limit 1.

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.