Does $\lim_{n\rightarrow \infty } {\left| a_n \right|} = {\left| L \right|} $ hold for$ L \neq 0$?

Question

I have this question:

Prove that $\large \lim_{n\rightarrow \infty } {a_n} = L \iff \lim_{n\rightarrow \infty} {\left| a_n \right|} = |L|$ when $L = 0.$

Does that hold for L $\neq 0$ ?

Well, I approached this problem by saying:

take $\large \varepsilon > 0$, then since $\large \ a_n \rightarrow L $, we have :

$\large \left | \left |a_n \left | - \right | L \right | \right | \leq \left| a_n - L \right| < \varepsilon $

therefore,

$\large \left | \left |a_n \left | - \right | L \right | \right | < \varepsilon$

and

$\large \left | a_n \left | \rightarrow \right | L \right | $

However, this turned out to be WRONG !

Any suggestions please

Answer 1

The statement is false: Take $a_n = (-1)^n$, and $L = 1$. Then $|a_n| = 1$ for every $n$, but $\lim_{n \to \infty} a_n$ doesn't exist.

Answer 2

This absolutely does not hold for $L\neq 0$.

Suppose that it did. Then if $\lvert a_n\rvert\rightarrow \lvert L\rvert$, then it must be true that $a_n\rightarrow L$.

However, note that $\lvert a_n\rvert\rightarrow\lvert L\rvert$ is equivalent to saying that $\lvert a_n\rvert\rightarrow \lvert-L\rvert$, since $\lvert-L\rvert=\lvert L\rvert$. But it then must be true that $a_n\rightarrow-L$ as $n\rightarrow\infty$.

However, by uniqueness of limits, a sequence cannot have two distinct limits... and if $L\neq 0$, $L$ and $-L$ are surely distinct.

Answer 3

If $L \neq 0$, then $\lim_n a_n = L $ implies $\lim_n |a_n| = |L| $ because $x \mapsto |x|$ is continuous ($||x|-|y|| \le |x-y|$).

However, if you take $a_n = (-1)^n$, then $a_n$ has no limit, but $|a_n| = 1$ for all $n$.