Limsup Squeeze Theorem

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

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

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

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

Therefore, we know $$0 \le \liminf_{n \to \infty} \left({y_n}\right) \le \limsup_{n \to \infty} \left({y_n}\right)$$, where $$\liminf$$ denotes the limit inferior.

So, $$\liminf_{n \to \infty} \left({y_n}\right) = \limsup_{n \to \infty} \left({y_n}\right) = 0$$, by the Squeeze Theorem.

Thus, $$\lim_{n \to \infty} \left({y_n}\right) = 0$$, but $$0 \le \left|{x_n}\right| \le y_n \implies \lim_{n \to \infty} \left|{x_n}\right| = 0$$

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

Then since $$-\left|{x_n}\right| \le x_n \le \left|{x_n}\right|$$, it follows by the Squeeze Theorem that $$\lim_{n \to \infty} x_n = 0$$.