Proof of Limit if and only if limit of absolute value is the same The question from my textbook is as follows:

Prove any sequence $\{a_n\}$ that $\lim_{n\to\infty} a_n = 0$ if and only if $\lim_{n\to\infty} |a_n| = 0$.

My attempt:
If $\lim_{n\to\infty} a_n = 0$, then for every $\epsilon > 0$, there exists an $n\in N$ such that $|an -0| = |a_n| = ||a_n|-0| < \epsilon$. Thus $\lim_{n\to\infty} |a_n| = 0$.
If $\lim_{n\to\infty} |a_n| = 0$, then for every $\epsilon > 0$, there exists an $n\in N$ such that $||a_n| - 0| = a_n = |a_n - 0| < \epsilon$. Thus $\lim_{n\to\infty} a_n = 0$.
I thought that my proof is a bit sketchy. Is my proof a valid proof? If not, how would I go about solving this?
 A: Your only problem is that you don't have the correct definition of limit.
$\lim_{n\rightarrow \infty} a_n = c$ means that for any $\epsilon$ there exists and $M \in \mathbb R$ so that $n > M; n \in \mathbb N$ will imply $|a_n - c| < \epsilon$.   
You stated something entirely different.  
Then as $|a_n - 0| = |a_n| = ||a_n| - 0|$ we have $|an - 0| < \epsilon \iff ||a_n| - 0|<\epsilon$ and so $\lim_{n\rightarrow \infty} a_n =0 \iff \lim_{n\rightarrow \infty} |a_n| =0$.
The part of your proof that felt sketchy was just fine.  But the part that felt just fine (the definition of limit) was wrong.
A: It doesn't look fine to me.
For example for the first part, you argued that
$$
\forall \epsilon >0, \exists n\in\mathbb{N} : ||a_n|-0|<\epsilon\tag{1}
$$
and concluded that
$$
\displaystyle\lim_{n\to\infty}|a_n|=0\tag{2}
$$
While $(1)$ is true, it is not equivalent to $(2)$. The meaning of $(2)$ is
$$
\forall \epsilon >0, \exists N\in\mathbb{N},\forall n\in\mathbb{N} : n>N \implies ||a_n|-0|<\epsilon\tag{3}
$$
Of course the same argument you gave will lead you to $(3)$.

Note: Define
  $$
a_n := \begin{cases}\frac{1}{n} & n \text{ even}\\1 & n \text{ odd}\end{cases}
$$
  Then the sequence $(a_n)$ satisfies $(1)$ but not $(2)$.

As for the second implication you tried to prove, you ran into the same problem and also the equality $||a_n| - 0| = a_n$ isn't true if $a_n<0$ (which can happen).

All you need in order to prove the result is to note (as you did) that
$$
|a_n-0| = ||a_n|-0|
$$
Then
$$
\lim_{n\to\infty} a_n=0\\
\Updownarrow \\
\forall \epsilon >0, \exists N\in\mathbb{N},\forall n\in\mathbb{N} : n>N \implies |a_n-0|<\epsilon \\
\Updownarrow \\
\forall \epsilon >0, \exists N\in\mathbb{N},\forall n\in\mathbb{N} : n>N \implies ||a_n|-0|<\epsilon \\
\Updownarrow \\
\lim_{n\to\infty} |a_n|=0
$$
A: It's not correct. You're emphasizing that there exists (at least) one $n$ that fits. But what you should say is those n large enough will fit.

$\epsilon$-$\delta$$(\epsilon$-$K)$ method is like a machine:
  For any given $\epsilon$, you put the $\epsilon$ into it, and get a corresponding $\delta$$(K)$.

In your proof, you don't take the advantage of the correspnding $K^\dagger$ from the definition of $\lim_{n\to\infty}(\bullet)$.
$^\dagger K\in \Bbb{Z}^+$.
Assume LHS is true: $$\forall \epsilon\gt 0, \exists K \in \Bbb{N}, \forall n\in \Bbb{N}: n\ge K\implies |a_n - 0|\lt\epsilon,$$
which implies the RHS is true, by the reasoning you given, if you choose the just corresponding K to limit your $n$. The reverse is similar.
