Recall that a sequence ($\displaystyle a_n$) converges to L if and only if

$\displaystyle

\forall \epsilon >0 \exists N \in \mathbb{N} \forall n \geq N |a_n - L|< \epsilon

$

Prove that if $\displaystyle a_n \rightarrow L$ then $\displaystyle |a_n| \rightarrow |L|$.

Attempt:

Suppose $\displaystyle a_n \rightarrow L$, then $\displaystyle \lim_{x-> \infty} a_n = L$

Then we know $\displaystyle -a_n \leq a_n \leq a_n$

The limit of the two outside terms is L, and hence the limit of $\displaystyle a_n$ is L by the squeezing theorem for sequences.

Conversely we can say: suppose $\displaystyle |a_n| \rightarrow L$ then

$\displaystyle -|a_n| \leq a_n \leq |a_n|$

etc...

But I doubt this is right or even the right approach. Can anyone show me a better and more rigorous proof?