Theoretical Limits

1. Jan 5, 2006

Orion1

My calculus book describes the following example:

Equation:
$$\lim_{x \rightarrow 0} \sin \frac{\pi}{x} = \text{does not exist}$$

However, my independent research indicates that it is actually:
Equation:
$$\lim_{x \rightarrow 0} \sin \frac{\pi}{x} = \text{Interval}(-1,1)$$

Is this second solution only theoretical, and not a real valid solution in a Calculus I course?

2. Jan 5, 2006

d_leet

That's the thing as x approcahes zero teh value of this function continually oscillates between -1 and 1 therefore the limit does not exist.

3. Jan 5, 2006

Pyrrhus

Like d_leet said for a limit to have a value L

$$\lim_{x \rightarrow a} f(x) = L$$

then

$$\lim_{x \rightarrow -a} f(x) = \lim_{x \rightarrow +a} f(x) = L$$

In the case of that function, it oscilates.

4. Jan 5, 2006

shmoe

I will hazard a guess that "independent research" means typing it into a program like Maple or Mathematica. You should have a look at the documentation provided by these programs as to what this actually means. From maple:

"If limit returns a numeric range it means that the value of the limiting expression is known to lie in that range for arguments restricted to some neighborhood of the limit point. It does not necessarily imply that the limiting expression is known to achieve every value infinitely often in this range."

5. Jan 5, 2006

bomba923

What 'research' led you to conclude that ??

Perhaps you meant to say:
$$\forall \varepsilon > 0,\bigcup\limits_{\left| x \right| < \varepsilon } {\left\{ {\sin \frac{\pi } {x}} \right\}} = \left[ { - 1,1} \right]$$
where $x \ne 0$ (an undefined element?)

Last edited: Jan 6, 2006
6. Jan 6, 2006

Orion1

I performed some calculations on this function and produced the following results:

$$\lim_{x \rightarrow 0^+} \sin \frac{\pi}{x} = 1$$
$$\lim_{x \rightarrow 0^-} \sin \frac{\pi}{x} = -1$$
$$\boxed{\lim_{x \rightarrow 0^{\pm}} \sin \frac{\pi}{x} = \pm 1}$$

Last edited: Jan 6, 2006
7. Jan 6, 2006

Tide

Perhaps you would be kind enough to show us your calcuations. :)

8. Jan 6, 2006

Hurkyl

Staff Emeritus
That is incorrect.

It is true that [-1, 1] is the set of limit points of this function as x goes to 0, but the limit of a function is defined to be its only limit point. Since this function does not have a single limit point (it has many), the limit does not exist.

9. Jan 6, 2006

HallsofIvy

Your independent research seems to have involved faulty arithmetic.

As Hurkyl said, [-1,1] is the set of limit points. I myself would have said "subsequential limits": given any number, X, between -1 and 1, it is possible to choose a sequence, {xn}, converging to 0, such that the sequence ${sin(\frac{\pi}{x_n})}$ converges to X. However, saying that is exactly saying that the limit of the function itself does not exist.

Last edited by a moderator: Jan 6, 2006
10. Jan 6, 2006

Orion1

Calculus Calculations...

Mathematica:
$$\text{In[1] := Limit} \left[ \sin \left[ \frac{\pi}{x} \right], x \rightarrow 0 \right]$$

$$\text{Out[1] = Interval[{-1,1}]}$$

$$\boxed{\forall \varepsilon > 0,\bigcup\limits_{\left| x \right| < \varepsilon } {\left\{ {\sin \frac{\pi }{x}} \right\}} = \left[ { - 1,1} \right] \; \; x \ne 0}$$

Basic Source Code:
Code (Text):

ST = 0.00001!

For X1 = 1 To 0 Step -ST
Y1 = Sin(PI / X1)
Next X1

For X2 = -1 To 0 Step ST
Y2 = Sin(PI / X2)
Next X2

Results:
$$\begin{array}{|c|c|} S_T & Y_1 \\ \hline 0.01 & 0.275 \\ 0.001 & 0.562 \\ 0.0001 & 0.832 \\ 0.00001 & 0.956 \\ \hline \end{array}$$

$$\begin{array}{|c|c|} S_T & Y_2 \\ \hline 0.01 & -0.275 \\ 0.001 & -0.562 \\ 0.0001 & -0.832 \\ 0.00001 & -0.956 \\ \hline \end{array}$$

$$\boxed{\lim_{x \rightarrow 0^{\pm}} \sin \frac{\pi}{x} = \pm 1}$$

$$\boxed{\lim_{x \rightarrow 0} \sin \frac{\pi}{x} = \text{does not exist}}$$

Last edited: Jan 6, 2006
11. Jan 7, 2006

Tide

Thanks, Orion.

Others have already explained the problem with Mathematica.

Your problem with BASIC is that none of your FOR loops ever hits the desired values of x1 = 0 or x2 = 0. The first clue is that your program didn't crash - i.e. you would have gotten a "division by zero" message. You are working with single precision numbers (which, by the way, according to the IEEE standard maintains only 6 digits of mantissa, i.e. significant digits).

By repeatedly subtracting ST from 1 (for small ST) the binary roundoff error never gets you to exactly zero so the compiler (interpreter?) uses the last value of x! which is nonnegative (in the case of X1 and conversely for X2). You just happened to get values that APPEAR to be getting closer to 1.0 but in reality that is completely fortuitous. Try a few smaller values for ST and your compiler will go bonkers (that's a technical term!) ;)

The correct answer to your original problem is that the limit simply does not exist - whether you approach it from the right or from the left. As others have already indicated, as you approach x = 0 the function will take on any of the values between -1 and +1 but the limit does not exist.

12. Jan 7, 2006

arildno

Orion:
In order to convince you of what others have said, consider a sequence of number $x_{k}=\frac{\pi}{\theta+2k\pi}[/tex], where [itex]\theta$ is some fixed number.

This sequence converges to 0 as k grows.
Form another sequence $y_{k}$ as follows:
$y_{k}=\sin(\frac{\pi}{x_{k}}[/tex] As is readily found, we have for all k, [itex]y_{k}=\sin\theta$.

Consider why this shows that the limit of your function cannot exist at 0, and why it proves that the set of limit points must be the interval [-1,1]

13. Jan 7, 2006

shmoe

One more thing, if you're looking at $$\lim_{x\rightarrow 0}f(x)$$ for example, it is not sufficient to look at f evaluated along some sequence approaching 0. e.g. if you punch f(1), f(.1), f(.01), f(.001), ..., f(10^(-n)), even if it "looks like" this sequence is converging to L say, you HAVE NOT just proven $$\lim_{x\rightarrow 0}f(x)=L$$. They like to put tables of these sorts of evaluations in first year calculus texts, but this is not at all a rigorous approach and can lead to terribly misleading and just plain wrong results.

14. Jan 8, 2006

Orion1

Limits...

I understand. However, what happens to this function when plotted by Mathematica at a domain of around +,- 5*10^-19?

The function itself appears to 'shut down' at around +,- 3.41*10^-19.

Is this a machine language or processor limitation?

Mathematica:
$$\text{Plot} \left[\sin \left[ \frac{\pi}{x} \right], {x,-1,1} \right]$$

$$\text{Plot} \left[\sin \left[ \frac{\pi}{x} \right], {x,-5*10^{-19}, 5*10^{-19}} \right]$$

The Mathematica plots from these commands are attached.

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Last edited: Jan 8, 2006
15. Jan 8, 2006

Zurtex

Well do remember that everything has its computational limits, though when I ran that on Mathematica it didn't result in the same problem.

16. Jan 9, 2006

Data

Mathematica is a computational tool, it usually can't help you deal with situations like this (in this case, a function that oscillates arbitrarily quickly by getting sufficiently close to zero) in a reasonable manner, because it is run on a machine in the physical, discrete, world. Frankly, in this case the software (or your processor) probably includes a feature specifically designed to stop your computer from generating an error!