### CSIR JUNE 2011 PART B QUESTION 31 SOLUTION (Limit points of $\{\frac{1}{n}sin\frac{1}{n}: n \in \Bbb N\}$)

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 $-1,0,1$.
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.
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...