# Squeeze Rule Proof variant

The usual proof for the squeeze rule for sequences is the following:

However I did a different one but it looks strange that it is normally not used as it's way shorter. First of all I write down the definitons:

$(a_n)$ converges to $L$ $$\forall \epsilon >0 \exists n_1 \in \mathbb{N}:|a_n-L|\leq \epsilon \forall n\in \mathbb{N}:n>n_1$$

$(c_n)$ converges to $L$: $$\forall \epsilon >0 \exists n_2 \in \mathbb{N}:|c_n-L|\leq \epsilon \forall n\in \mathbb{N}:n>n_2$$

We have $a_n \leq b_n \leq c_n$ $\forall n\in \mathbb{N}$. Hence $$a_n-L\leq b_n-L\leq c_n - L$$ $\forall n\in \mathbb{N}$, hence $$a_n-L\leq b_n-L\leq c_n - L \leq |c_n-L|\leq \epsilon$$ which holds $\forall n > n_2$. But we also have $$-c_n+L\leq -b_n+L\leq -a_n + L$$ which again gives $$-c_n+L\leq -b_n+L\leq -a_n + L\leq |a_n-L| < \epsilon$$ which holds $\forall n > n_1$.

Hence just take $n_0:=\max\{n_1,n_2\}$ and $\forall n>n_0$ we have that $|b_n-L| \leq \epsilon$ $\forall \epsilon >0$

Does this even work?

• Yes, this works. – KittyL Dec 30 '16 at 18:56