What happens to $f(x) = \left\lceil \dfrac{x}{a} \right\rceil \cdot a$ as $x \rightarrow \infty$? Consider the function $f(x) = \left\lceil \dfrac{x}{a} \right\rceil \cdot a~~~$ 
where $a \in R$ and $a \neq 0 $. Now let us say we are interested in the behavior of $f(x)$ as $x \rightarrow \infty$. It seems like $f(x) \sim x$, but I'm trying to come up with a formal proof. For an illustration, refer here.
 A: From the rule
$$t\le\lceil t\rceil<t+1$$ you deduce
$$1\le\frac{f(x)}x<1+\frac ax$$ and this squeezes to $1$.
The answer is similar for $a<0$.
A: $\frac xa \le \lceil \frac xa \rceil < \frac xa + 1$
$x \le \lceil \frac xa \rceil*a < x + a$.
And this will be true for all $x$. But we can never say the difference $\lceil \frac xa \rceil*a$ and $x$ ever decreases absolutely.
I think what you are trying to say is that $f(x)\approx x$ with an absolute margin of error of $a$.  But as $x$ gets very large that margin of error becomes less significant.
$1= \lim \frac xx \le \lim \frac {\lceil \frac xa \rceil*a}x < \lim \frac {x+a}x = 1$
So $\lim_{x\to \infty}\frac {\lceil \frac xa \rceil*a}x=1$
And $f(x)\approx x$ as $x$ gets large.  But it doesn't make sense to talk of the limit of $f(x)$ approaching $g(x)$.  As $x\to \infty$ it makes no sense to talk of a $x$ being a variable in $g(x)$.  Also $|f(x) -g(x)|\not \to 0$.
For any $a - \epsilon$ where $0< \epsilon < a$ we can always find large values of $x$ where $|\lceil \frac xa \rceil*a -x | >a-\epsilon$.
