The set $\{\frac{1}{n}sin\frac{1}{n}: n \in \Bbb N\}$ has 1) one limit point and it is 0,

2) one limit point and it is $1$,

3) one limit point and it is $-1$,

4) three limit points and these are $-1,0,1$.

We claim that the only limit point of the given set is zero and hence the

Since the sequence $\{\frac{1}{n}sin\frac{1}{n}\}$ converges to zero, we have zero is a limit of the given set. Let $a$ be a limit point of this set, then there will be a sequence in this set which converges to $a$. But any sequence in the given set is a subsequence of the sequence $\{\frac{1}{n}sin\frac{1}{n}\}$ and by the above argument, this sequence should converge to zero. Since the limit of a sequence is unique we have $a=0$ and $0$ is the only limit point.

2) one limit point and it is $1$,

3) one limit point and it is $-1$,

4) three limit points and these are $-1,0,1$.

**Solution**: Since $|sin\frac{1}{n}| \le 1$, we have the sequence $\{sin\frac{1}{n}\}$ is bounded. This shows that the sequence $\{\frac{1}{n}sin\frac{1}{n}\}$ converges to zero. Hence any the subsequences of this sequence are also converging to zero.We claim that the only limit point of the given set is zero and hence the

**correct option is (1)**.Since the sequence $\{\frac{1}{n}sin\frac{1}{n}\}$ converges to zero, we have zero is a limit of the given set. Let $a$ be a limit point of this set, then there will be a sequence in this set which converges to $a$. But any sequence in the given set is a subsequence of the sequence $\{\frac{1}{n}sin\frac{1}{n}\}$ and by the above argument, this sequence should converge to zero. Since the limit of a sequence is unique we have $a=0$ and $0$ is the only limit point.

**Share to your groups:**FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS AND COMMENTS BELOW IN THE COMMENTS SECTION. ALSO, SUGGEST PROBLEMS TO SOLVE.

## No comments:

## Post a Comment