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


For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template and an example, see this.

Problem Discussion

We’ll follow a similar approach to proving the example in here. We want to show that \(\lim (y_n) = l\). So we want to find \(N \geq n\) such that for any \(\epsilon > 0\),

$$ \begin{align*} \lvert y_n - l \rvert < \epsilon. \end{align*} $$

But we know that since \((x_n)\) converges to \(l\), then there must exist an \(n \geq N_1\) such that

$$ \begin{align*} \lvert y_n - l \rvert < \epsilon \quad \text{whenever $n \geq N_1$.} \end{align*} $$

Similarly there must exist some \(n \geq N_2\) such that

$$ \begin{align*} \lvert z_n - l \rvert < \epsilon \quad \text{whenever $n \geq N_2$.} \end{align*} $$

Let’s get rid of the absolute value for both inequalities and then add \(l\) to all sides.

$$ \begin{align*} -\epsilon &< y_n - l < \epsilon \\ l - \epsilon &< y_n < l + \epsilon \\ \end{align*} $$

We can derive the exact result for \(z_n\) as well.

$$ \begin{align*} l - \epsilon &< z_n < l + \epsilon \\ \end{align*} $$

Now from these two inequalities, notice that \(x_n \leq y_n\) and so we can write

$$ \begin{align*} l - \epsilon < x_n \leq y_n \quad \text{whenever $n \geq N_1$.} \end{align*} $$

and we also have

$$ \begin{align*} y_n \leq z_n < l + \epsilon \quad \text{whenever $n \geq N_2$.} \end{align*} $$

Combining both inequalities will get us what we want! let’s expand on this in the formal proof.

Formal Proof

To show that \(\lim(y_n) = l\), let \(\epsilon > 0\) be arbitrary. We want to prove that there exists some \(n \geq N\) such that,

$$ \begin{align*} \lvert y_n - l \rvert < \epsilon \quad \text{whenever $n \geq N$}. \end{align*} $$

To do so, we are given that \(\lim x_n = l\). This means that there exists some \(n \geq N_1\) such that

$$ \begin{align*} \lvert x_n - l \rvert < \epsilon \quad \text{whenever $n \geq N_1$}. \end{align*} $$

Similarly, we are given that

$$ \begin{align*} \lvert z_n - l \rvert < \epsilon \quad \text{whenever $n \geq N_2$}. \end{align*} $$

We can re-write the inequalities such that

$$ \begin{align*} l - \epsilon < x_n < l + \epsilon \quad \text{whenever $n \geq N_1$} \\ l - \epsilon < z_n < l + \epsilon \quad \text{whenever $n \geq N_2$}. \end{align*} $$

Now assume that \(n \geq \max{(N_1, N_2)}\), using the above inequalities and the fact that \(x_n \leq y_n \leq z_n\), we can see that

$$ \begin{align*} l - \epsilon < x_n \leq y_n \\ y_n \leq z_n < l + \epsilon. \end{align*} $$

Combining both inequalities, we get

$$ \begin{align*} l - \epsilon < x_n \leq y_n \leq z_n < l + \epsilon \\ l - \epsilon < y_n < l + \epsilon \\ - \epsilon < y_n - l < \epsilon \\ |y_n - l| < \epsilon. \end{align*} $$

From this we can conclude that for any \(\epsilon > 0\) there exists an \(N \geq n\) such that \(|y_n - l| < \epsilon\). Therefore, \(\lim y_n = l\) as we wanted to show. \(\blacksquare\)

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