Which is the correct answer from the following options- Let $(x_n)$ be a sequence of real numbers such  that the subsequences $(x_{2n})$  &$(x_{3n})$ converge to limits $K$ and $L$ respectively.Then-


*

*$(x_{n})$ always converges.

*If  $K=L$,then $(x_n)$ converges.

*$(x_n)$ may not converge,but $K=L$.

*it is possible to have $K \neq L.$


This question has been asked in ISI-2017 entrance exam.Last two options are correct but i'm not able to justify it.Moreover,i'm not getting how to discard first two options.
Also please suggest me some reference from where i can get these kind of problems.
Any kind of hint or help or suggestions are heartly welcome!!
 A: Hint:
You can discard the first two options by looking at $x_p$ for prime values of $p$

Also, it is not true that the last two options are correct, since the last two options clearly contradict each other (one claiming $K=L$ is always true, the other saying $K\neq L$ is possible).
In fact, only the third option is true, and proving $K=L$ is most efficient by looking at the limit of $x_{6n}$ as $n\to\infty$.
A: $(x_{2n})$ and $(x_{3n})$ share a common infinite subsequence, namely $(x_{6n})$.

Any infinite subsequence of a convergent sequence converges, and moreover, converges to the same limit as the full sequence.

It follows that $K=L$.

To see that the sequence $(x_n)$ need not converge, consider the sequence defined by
$$x_n = 
\begin{cases}
1 &\text{if}\;n\;\text{is relatively prime to $6$}\\[4pt]
0 &\text{otherwise}\\[4pt]
\end{cases}
$$ 
A: Let
\begin{equation*}
x_n = \begin{cases}
0 &\text{if } 2 \text{ or } 3 \text{ divides } n \\
n &\text{ otherwise }
\end{cases}
\end{equation*}
For this sequence, we have $K = L = 0$, but $(x_n)_{n \in \mathbb{N}}$ is not bounded, and therefore does not converge. So both options 1 and 2 are false.
Suppose that $K \neq L$, and let $\varepsilon > 0$ such that $\varepsilon < ( K - L ) / 3$. From the convergence of $(x_{2n})$, there exists $N_1 \in \mathbb{N}$ such that $| L - x_{2n} | < \varepsilon$ for all $n \geq N_1$. Similarly, from the convergence of $(x_{3n})$, there exists $N_2 \in \mathbb{N}$ such that $| K - x_{3n} | < \varepsilon$ for all $n \geq N_2$. Let $N = \max \{ N_1, N_2 \}$. The term $x_{6N}$ appears both in $(x_{2n})_{n \in \mathbb{N}}$ (at index $3N$) and in $(x_{3n})_{n \in \mathbb{N}}$ (at index $2N$), and both $2N$ and $3N$ are greater than $N$. Therefore we must have $| x_{6N} - L | < \varepsilon$ and $| x_{6N} - K | < \varepsilon$. This leads to $| K - L | \leq | K - x_{6N} | + | x_{6N} - L | < 2 \varepsilon \leq 2 / 3 \cdot ( K - L )$, which is absurd. So we must have $K = L$. Option 3 is then true, and option 4 false.
