If $\lim_{n \rightarrow \infty}x_n=\lim_{n \rightarrow \infty}y_n=L$ then the sequence $\{x_1,y_1,x_2,y_2,...\}$ tends to $L$

Question

Let $\lim_{n\rightarrow\infty} x_n = L$ and $\lim_{n\rightarrow\infty} y_n = L$. Then show that the using definition of convergence that sequence $z_n = (z_1, z_2, z_3, z_4,...) = (x_1, y_1, x_2, y_2, ....)$ converges to $L$.

My attempt:

Clearly, $|x_n-L|<\epsilon$ $\forall n>k_1$ and $|y_n-L|<\epsilon$ $\forall n>k_2$.

$z_{n}$ can be separated into the cases $z_{2m}$ and $z_{2m+1}$ where $m\in\mathbb{N}$.

We observe, $z_{2m}=y_m$ and $z_{2m+1}=x_{m+1}$ . Thus, both the odd and even terms of $z_n$ converge separately to the same limit $L$ since $x_n$ and $y_n$ converge to $L$.

Now, the last part of the logic is not proper and mathematically rigorous. I need help in stating that more formally.

Answer 1

Fix $\epsilon > 0$, and suppose for $n > M_1$, we have $|x_n - L| < \epsilon$ (possible since $x_n \to L$). Similarly, suppose for $n > M_2$, $|y_n - L| < \epsilon$. Choose $M = 2\max\{M_1, M_2\}$, and for $n > M$ we have $|z_n - L| < \epsilon$.