Limit properties as $x\to \infty$ for functions 1) Let $q\in(0,1)$ is fixed and $L$ is a finite value. Is it possible to say if
$\lim_{x\to\infty}f(qx)=L$ then $\lim_{x\to\infty}f(x)=L.$ 
2) And i also stack in if for all $\epsilon>0$ is it possible to say if
$L-\epsilon\leq f(\frac{x}{q})$ and $f(qx)\leq L+\epsilon$ then $\lim_{x\to\infty}f(x)=L.$
 A: Ad. 1:
Yes, it is. Read the definition of limit.
$$(\forall x>M_{\epsilon})(|f(x) - L| < \epsilon) \Longrightarrow
(qx > M_{\epsilon} \Rightarrow (|f(qx) - L| < \epsilon)) \Longrightarrow\\
(\forall x > \frac{M_{\epsilon}}{q})
(|f(qx) - L| < \epsilon)$$
Ad. 2: Yes it is true. As above, in abbreviated form. (In fact, it's hard to understand question. At first time, I was thinking about "for all", now I think it's "for big enough".)
$$(\forall x \in \mathbb{R})\left(x > M_{\epsilon} \Rightarrow L-\epsilon \leq f\left(\frac{x}{q}\right)\right) \Longrightarrow (\forall x > \frac{M_{\epsilon}}{q})(L-\epsilon \leq f(x))$$
Similarly you can receive $(\forall x > M_{\epsilon})(x \leq L+\epsilon)$.
$$(\forall x \in \mathbb{R})\left(x > M_{\epsilon} \Rightarrow
L+\epsilon \geq f\left(qx\right)\right) \Longrightarrow
(\forall x > \frac{M_{\epsilon}}{q}>M_{\epsilon}q)(L+\epsilon \geq f(x))$$
This gives us almost final conclusion. $$(\forall x > \frac{M_{\epsilon}}{q})(L-\epsilon \leq f(x) \leq L+\epsilon)$$
So we can easily receive $\lim_{x\to\infty} f(x) = L$.
$$\begin{split}
(\forall x > M > \frac{M_{\epsilon}}{q} )(L-\epsilon \leq f(x) \leq L+\epsilon) &
&\Longleftrightarrow
(\forall x > M > \frac{M_{\epsilon}}{q} )(\left|f(x)-L\right|\leq\epsilon)\\
\Longleftrightarrow \lim_{x\to\infty}f(x) = L
\end{split}$$
$\mathscr{Q.E.D.}$
