4

A Problem : \begin{equation}\lim_{x\to 0} \frac{\sin x}{x}\end{equation} results in the solution : $1$

But the same function enclosed in a greatest integer function results in a $0$

\begin{equation}\lim_{x\to 0} \left\lfloor{\frac{\sin x }{x}}\right\rfloor\end{equation}

Why?

My thoughts:
[The value of the first function tends to 1 because of the expansion :

$$\frac{\sin\left( x \right)}{x}\approx\frac{ x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots}{x}$$ $$\approx 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots$$ and putting zero in the function results in 1

but applying the greatest integer function to the same will result in a zero as whenever the value of the result is taken it will be slightly less than one because of all the subtractions involved in the expansion.]

mich95
  • 8,713
The-Ever-Kid
  • 560
  • 1
  • 4
  • 13
  • 4
    Isn't it just as simple as noting that $|\sin x|<|x|$ when $x\neq 0$? This has a relatively simple geometric meaning. – Thomas Andrews May 16 '12 at 17:03
  • @ThomasAndrews Meaning sinx will always be less than x for x$\neq$0 so the result will be a fraction hence the result for the GIF is zero the why is the result one without it? – The-Ever-Kid May 16 '12 at 17:08
  • 2
    Your thoughts are essentially correct. It is possible for a function or sequence to tend to $1$ while always being less than $1$, take the sequence $1-\frac{1}{n}$ for a simpler example. What you have proved is that the greatest integer function does not commute with taking limits. This should not be too surprising, as the greatest integer function is not continuous. – mdp May 16 '12 at 17:12
  • @MattPressland "the greatest integer function does not commute with taking limits" meaning? – The-Ever-Kid May 16 '12 at 17:14
  • 2
    Just wondering, is there any particular reason for writing $\approx$ instead of $=$ when using the power series? – Antonio Vargas May 16 '12 at 17:15
  • For the geometric interpretation mentioned by Thomas, see this question: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?. Another question related to inequality $|\sin x|<|x|$ is this: How to strictly prove $\sin <x$ for $0<x<\frac{\pi}{2}$. – Martin Sleziak May 16 '12 at 17:16
  • 2
    @The-Ever-Kid Fair question - it means that it matters whether you take the limit and then apply the greatest integer function, or do it the other way round. The reason why this happens is that the greatest integer function is not continuous (this is something worth finding out about if you aren't familiar with it). – mdp May 16 '12 at 17:20
  • 1
    Or in the same spirit: $$\lim_{x \to 0} \lfloor 1-x \rfloor$$ – azdahak May 16 '12 at 20:54
  • @MattPressland: Perhaps you'd like to post an answer summarizing your comments? – Zev Chonoles May 17 '12 at 01:32
  • 1
    @ZevChonoles Sure thing, I'll do that now. Sorry it took so long, it was night in my time zone. – mdp May 17 '12 at 09:43

1 Answers1

2

The problem that is ocurring in your example is that the greatest integer function is not continuous. One consequence of this is that in general the limits $\lfloor \lim_{x\to c}f(x)\rfloor$ and $\lim_{x\to c}\lfloor f(x)\rfloor$ will not be equal to each other. This happens because the greatest integer function changes value abrubtly as you move along $\mathbb{R}$; every time you hit an integer, the value jumps up by $1$. So if, as $x\to c$, the function $f(x)$ approaches an integer $n$ from below, as in the case of $\frac{\sin{x}}{x}$ as $x\to 0$, the greatest integer function will (for $x$ close enough to $c$) return $n-1$, as $f(x)$ eventually exceeds $n-1$ as $x\to c$, but it never gets as high as $n$. On the other hand, if you take the limit first, you "attain" $n$, and then the greatest integer function returns $n$.

In general, $g(\lim_{x\to c}f(x))=\lim_{x\to c}(g(f(x))$ for all $c$ and $f$ such that $\lim_{x\to c}f(x)$ actually exists if and only if the function $g$ is continuous. This has several equivalent definitions, but can be thought of roughly as meaning the graph of $f$ has no sudden jumps (and doesn't oscillate too fast, like $\sin\frac{1}{x}$).

There are lots of big theorems about when you can swap the order of taking a limit and applying some other function, and when you can change the order of two limits etc. This turns up quite a lot with theorems about differentiation, integration and series, and mixtures of the three, as all are defined in terms of limits.

mdp
  • 14,671
  • 1
  • 39
  • 63