Squeeze Theorem

analysis

Definition

Squeeze Theorem

Let $(a_n{)}_{n\ge 0}$ and $(b_n{)}_{n\ge 0}$ be convergent sequences with identical limits ${\lim }_{n\to \infty }{a}_n={\lim }_{n\to \infty }{b}_n=L$. Let $(c_n{)}_{n\ge 0}$ be a sequence such that:

$$a_n\le c_n\le b_n$$

for almost all $n\in \mathbb{N}$. Then $(c_n)$ is also convergent, and its limit is equal to $L$:

$$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{c}_n=\underset{n\to \infty }{\lim }{b}_n=L$$

Function Version

The theorem applies analogously to limits of functions:
If $f(x)\le g(x)\le h(x)$ for all $x$ in a neighbourhood of $x_0$(except possibly at $x_0$), and ${\lim }_{x\to x_0}f(x)={\lim }_{x\to x_0}h(x)=L$, then ${\lim }_{x\to x_0}g(x)=L$.

Proof

Squeeze Theorem

Let $(a_n{)}_{n\ge 0}$ and $(b_n{)}_{n\ge 0}$ be convergent sequences with identical limits $\ell ={\lim }_{n\to \infty }{a}_n={\lim }_{n\to \infty }{b}_n$. Let $(c_n{)}_{n\ge 0}$ be a sequence with $a_n\le c_n\le b_n$.

For any $\epsilon >0$, by the definition of convergence:

1. There exists $N_1$ such that for all $n>N_1$,$|a_n-\ell |<\epsilon$, which implies $\ell -\epsilon <a_n<\ell +\epsilon$.
2. There exists $N_2$ such that for all $n>N_2$,$|b_n-\ell |<\epsilon$, which implies $\ell -\epsilon <b_n<\ell +\epsilon$.

Let $N=\max (N_1,N_2)$. Then for all $n>N$:

$$\ell -\epsilon <a_n\le c_n\le b_n<\ell +\epsilon$$

Thus, $\ell -\epsilon <c_n<\ell +\epsilon$, which means $|c_n-\ell |<\epsilon$. This proves that ${\lim }_{n\to \infty }{c}_n=\ell$.

$\square$