Show that $x_n \leq y_n \leq z_n$ for all $n \in \mathbb{N}$ and if $\lim{x_n} = \lim{z_n} = l$, then $\lim{y_n} = l$ as well.

Observe that $y_n \leq z_n$ and so $y_n - l \leq z_n - l \leq |z_n-l|$ and we know that for certain $N_1$, $n> N_1$ will imply $|z_n -l| < \epsilon$.

Similarly, because $y_n \geq x_n$ we can write $l-y_n \leq l-x_n \leq |l-x_n| = |x_n-l|$ and, again, we know that for certain $N_2$ $n>N_2$ will imply $|x_n-l| < \epsilon$.

Now, take N to be max($N_1$,$N_2$).

Is this proof idea correct?


Your idea is quite good, but to make the proof complete, I'd say you should explicitly argue that $|y_n-l|$ is equal to either $y_n-l$, in which case $|y_n-l|\le|z_n-l|$, or $l-y_n$, in which case $|y_n-l|\le|x_n-l|$, so that in either case $|y_n-l|\le\epsilon$ if $n\gt N=\max(N_1,N_2)$.

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