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I'm stumped by the following question. Can anyone help me out. Clearly the answer is l but I can't work out how to prove this

Suppose that $\{x_n\}$ is a sequence converging to a limit $l$. Define a new sequence $\{y_n\}$ by setting $y_n := x_{2n}$ for all $n \geq 1$. Determine the limit of $y_n$, giving a rigorous proof.

Thanks

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marked as duplicate by user159517, max_zorn, Delta-u, rtybase, Cesareo Feb 25 at 0:26

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$\{y_n\}$ is converging to $l$ if for every $ \varepsilon > 0$ there exists $N \in \mathbb{N}$ such that $|y_n-l| < \varepsilon$ for all $n \geq N$.
Now let $\varepsilon >0$, then there exists $M \in \mathbb{N}$ such that $|x_{2n}-l| <\varepsilon$ for all $n \geq \frac{M}{2}$ (since $\{x_n\}$ is converging to $l$). So we can choose $N:=\lfloor\frac{M}{2}\rfloor$ and we get $|x_{2n}-l|=|y_n-l| < \varepsilon$ for all $n \geq N$. Hence $\lim_{n \to \infty} y_n=l$.

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