Limsup Squeeze Theorem

Theorem

Let $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ be sequences in $\R$.

If:

• $\forall n \ge n_0: \left|{x_n}\right| \le y_n$
• $\displaystyle \limsup_{n \to \infty} \left({y_n}\right) = 0$, where $\limsup$ denotes the limit superior

then:

$\displaystyle \lim_{n \to \infty} x_n = 0$

Direct Proof

Since $\left|{x_n}\right| \ge 0$, we have that $y_n \ge 0$.

Therefore, we know:

$\displaystyle 0 \le \liminf_{n \to \infty} \left({y_n}\right) \le \limsup_{n \to \infty} \left({y_n}\right)$

where $\liminf$ denotes the limit inferior.

So:

$\displaystyle \liminf_{n \to \infty} \left({y_n}\right) = \limsup_{n \to \infty} \left({y_n}\right) = 0$

by the Squeeze Theorem.

Thus:

$\displaystyle \lim_{n \to \infty} \left({y_n}\right) = 0$

but:

$\displaystyle 0 \le \left|{x_n}\right| \le y_n \implies \lim_{n \to \infty} \left|{x_n}\right| = 0$

Therefore:

$\displaystyle \lim_{n \to \infty} \left({-\left|{x_n}\right|}\right) = 0$

Then since $-\left|{x_n}\right| \le x_n \le \left|{x_n}\right|$, it follows by the Squeeze Theorem that:

$\displaystyle \lim_{n \to \infty} x_n = 0$

$\blacksquare$