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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.

V-Red
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    Start with $\frac{x}{a}\leq \left \lceil \frac{x}{a} \right \rceil \leq \frac{x}{a}+1$ – rtybase Sep 03 '19 at 20:23
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    Doesn't "converge to $x$" simply mean the limit diverges? – Andrew Chin Sep 03 '19 at 20:26
  • @AndrewChin What I'm trying to say is, for the case when x>>a, what is f(x)? – V-Red Sep 03 '19 at 20:27
  • It is a step function with values in ${na : n\in\mathbb N}$. – amsmath Sep 03 '19 at 20:32
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    The statemeht that $\lim_{x\to\infty} f(x)$ converges to $x$ is non-sensical. There are other ways to write it, though. It is not true that $f(x)-x\to 0.$ However, it is true that $\lim_{x\to\infty} f(x)/x\to 1$ which is often written as $f(x)\sim x.$ – Thomas Andrews Sep 03 '19 at 20:51
  • @Thomas Andrews I've corrected the statement. – V-Red Sep 03 '19 at 23:03
  • You can always get $f(x) - x$ to be almost as large a $a$ by for any $M$ as large as you like and $n$ an integer $n>M$, taking $x= n*a + \epsilon$ for a very tiny $\epsilon$. Then $\frac xa=n+a \epsilon$ and $[\frac xa]=n+ 1$ and $[\frac xa]a=na + a = x-\epsilon +a$. So $f(x)-x=a-\epsilon$. ... the thing is the ratio of $a$ to $x$ because less significant for large values of $x$. – fleablood Sep 03 '19 at 23:52

2 Answers2

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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 $a<0$.

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$\frac xa \le \lceil \frac xa \rceil < \frac xa + 1$

$x \le \lceil \frac xa \rceil*a < x + a$.

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 $|\lceil \frac xa \rceil*a -x | >a-\epsilon$.

fleablood
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