Limsup Squeeze Theorem

Theorem
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.

Let:
 * $(1): \quad \forall n \ge n_0: \size {x_n} \le y_n$
 * $(2): \quad \ds \limsup_{n \mathop \to \infty} \paren {y_n} = 0$, where $\limsup$ denotes the limit superior.

Then:
 * $\ds \lim_{n \mathop \to \infty} x_n = 0$

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

Therefore, we know:
 * $\ds 0 \le \liminf_{n \mathop \to \infty} \paren {y_n} \le \limsup_{n \mathop \to \infty} \paren {y_n}$

where $\liminf$ denotes the limit inferior.

So:
 * $\ds \liminf_{n \mathop \to \infty} \paren {y_n} = \limsup_{n \mathop \to \infty} \paren {y_n} = 0$

by the Squeeze Theorem.

Thus:
 * $\ds \lim_{n \mathop \to \infty} \paren {y_n} = 0$

but:
 * $\ds 0 \le \size {x_n} \le y_n \implies \lim_{n \mathop \to \infty} \size {x_n} = 0$

Therefore:
 * $\ds \lim_{n \mathop \to \infty} \paren {-\size {x_n} } = 0$

Then since $-\size {x_n} \le x_n \le \size {x_n}$, it follows by the Squeeze Theorem that:
 * $\ds \lim_{n \mathop \to \infty} x_n = 0$