Existence finite limit implies bounded sequence Consider a sequence of real numbers $\{a_n\}_n$. I have read in several sources that
(*) If $\exists$ $\lim_{n\rightarrow \infty}a_n=L$ with $-\infty<L<\infty$ then $\{a_n\}_n$ is bounded i.e. $\exists$ $0<M<\infty$ s.t. $|a_n|\leq M$ $\forall n$
Isn't the correct statement 
(**) If $\exists$ $\lim_{n\rightarrow \infty}a_n=L$ with $-\infty<L<\infty$ then $\{a_n\}_n$ is bounded above and/or below? 
? 
Why? 
Do similar conclusions hold for a function $f:\mathbb{R}\rightarrow\mathbb{R}$ with $\lim_{x\rightarrow \infty}f(x)=L$ and $-\infty<L<\infty$?
 A: The first one is the correct one. By definition, $(a_n)_n$ converges to $L\in\mathbb{R}$ means
$$
\forall \varepsilon > 0 \exists N_\varepsilon \geq 0 \text{ s.t. } \forall n \geq N, \ \lvert a_n - L \rvert \leq \varepsilon.
$$
Take $\varepsilon = 1$, for instance:
$$
\exists N_1 \geq 0 \text{ s.t. } \forall n \geq N_1, \ \lvert a_n - L \rvert \leq 1.
$$
Now, by definition of the absolution values, this means that for all $n \geq N_1$
$$
L-1 \leq a_n \leq L+1.
$$
This implies that $(a_n)_n$ is bounded by $\ell=\max(\lvert L-1\rvert,\lvert L+1\rvert) $, except for maybe the first few (at most $N_1-1$) terms. But there are a finite number of them, so you are done: letting $M=\max(\ell, \lvert a_1\rvert,\dots, \lvert a_{N_1-1}\rvert)$, you then get that $\lvert a_n\rvert \leq M$ for all $n\geq 1$.
Edit: you can mimic this proof for continuous functions $f\colon\mathbb{R}\to\mathbb{R}$ having finite limits at $\pm\infty$. (the key is that at some point, you'll need to consider $\sup_{[-A,A]} \lvert f\rvert$ to conclude, and continuity ensures that this is finite.)
A: If $\{a_n\}_n\subset \Bbb R$ and $\lim_{n\rightarrow \infty}a_n=L\in \Bbb R=(-\infty,\infty)$, then
$$\exists N>0 \qquad\text{such that}\qquad |a_n-L|<1 \quad \forall n\geq N$$
this implies that
$$ -1<a_n-L<1 \implies |a_n|<\max\{|L+1|,|L-1|\} \qquad \forall n\geq N$$
So you can choose
$$M= \max\{|a_0|,\ldots,|a_N|,|L-1|,|L+1|\}$$
