Exercise 2.3.3

Question

[Squeeze Theorem]
  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.

Ulisse Mini · Accepted Answer

Let $\epsilon > 0$, set $N$ so that $|x_n - l| < \epsilon/4$ and $|z_n - l| < \epsilon/4$. Use the triangle inequality to see $|x_n - z_n| < |x_n-l|+|l-z_n|<\epsilon/2$.
  Note that since $x_n \leq y_n \leq z_n$, $|y_n - x_n| = y_n - x_n \leq z_n - x_n = |z_n - x_n|$.
  Apply the triangle inequality again to get
  $$|y_n - l| \le |y_n - x_n| + |x_n - l| \leq |z_n - x_n| + |x_n - l|< \epsilon/2 + \epsilon/4 < \epsilon$$

Exercise 2.3.3

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Exercise 2.3.3

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