# Find limit of a sequence $Y_n$ when $Y_n = X_{2n}$ and limit $X_n=l$ [duplicate]

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

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

$$\{y_n\}$$ 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 $$\lim_{n \to \infty} y_n=l$$.