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Trying to construct an example for a Business Calculus class (meaning trig functions are not necessary for the curriculum). However, I want to touch on the limit problem involved with the $\sin(1/x)$ function.

I am sure there is a simple function, or there isn't... But would love some insight.

I also understand that the functions that satisfy this condition are maybe way outside the scope of the course. I'm just looking for different "flavors" of showing limits that don't exist besides just showing the limit from the left and the limit from the right does not exists.

michael mccabe
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  • Can you clarify which limit problem you are referring to? – abiessu Jul 19 '21 at 17:14
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    Do you have a strict criterion on 'acts like' ? – user2628206 Jul 19 '21 at 17:14
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    The function $\sin(1/x)$ oscillates infinitely many times in the neighborhood of zero, something no polynomial or rational function can do. It might be worth revising your Question with particular traits you need to fulfill. – hardmath Jul 19 '21 at 17:17
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    I think your task is hopeless. Just give them $\sin\frac{1}{x}$. It won't kill them. – TonyK Jul 19 '21 at 17:17
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    you need a function with an infinite amount of roots. The simplest is probably the sin – user619894 Jul 19 '21 at 17:18
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    Do you mean "algebraic function" or "rational function"? I would think that the full generality of an algebraic function would be harder for Business Calculus students to understand than understanding $\sin(1/x)$. – Michael Seifert Jul 19 '21 at 17:21
  • Infinitely many roots near zero would be nice. But more specifically, looking for a function that has a limit that doesn't exist and isn't because the limit from the left doesn't equal the limit from the right.

    I want to explore different "flavors" of showing limits don't exists using numerical approaches.

     – michael mccabe Jul 19 '21 at 17:30
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    Just replace the $\sin(1/x)$ example with a piecewise linear graph whose line segments go up and down at ever larger slopes (positively and negatively) as $x \to 0$ from both directions. – KCd Jul 19 '21 at 17:43
  • Just draw an appropriate infinitely oscillating saw-tooth function on the blackboard (or whatever you use in lectures), or draw what $\sin (1/x)$ looks like. You seem to be under the impressing that you have to have a formula for the function, which will likely reinforce false concepts they may have about functions. Incidentally, obviously you're not going to get an algebraic function, as all algebraic functions (even implicitly defined) have finitely many turning points (even for the entire real line, much less in a bounded interval), for reasons you should know. – Dave L. Renfro Jul 20 '21 at 19:52

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As you note, this is really outside the scope of a business calculus syllabus. I might argue that anything more than a very informal discussion of limits is too.

In any case I think your business calculus students could profit from understanding that functions need not come from formulas. You can convey lots of the meaning and usefulness of calculus just with sketches of graphs. For this example you could sketch the graph near the origin at large magnification to show the infinitely many oscillations. If you draw the oscillations between the lines $y = \pm x$ you can get continuity. Between $y = \pm x^2$ you get differentiablity too.

Ethan Bolker
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  • (Lightbulb) Random function? I usually show the $\sin(1/x)$ as a different flavor of problematic functions that are hit with the limit. But I am searching for something that a Business Calculus student might relate to. Again, it may not exists, but why not ask... – michael mccabe Jul 19 '21 at 17:46
  • (+1) I didn't see your answer until I had posted my comment. For some reason when I scrolled down to see if there were answers (I have the view set at a moderately high magnification) I skipped past your answer, which says roughly what I said (with a little extra showing how to include various smoothness conditions -- regarding which, see this answer). – Dave L. Renfro Jul 20 '21 at 19:57
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Consider applying the fractional part function to $1/x^2$ or something similar. This would be an even function, so the behavior from the left is the same as the behavior from the right of zero, but neither limit from above or below exists because of oscillations.

Note that defining the fractional part function on negative numbers is done differently by various authors, so that's another reason to introduce $1/x^2$ and avoid that ambiguity.

I wouldn't call this an algebraic function, although the notions of integer part and fractional part of positive reals numbers should be pretty intuitive for your "Business Calculus" students. The fractional part function is not continuous, so it is scarcely surprising when limits involving it fail to exist.

hardmath
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I endorse Ethan's answer: define the function by drawing a graph. You know that it's $ x \mapsto \sin ( 1 / x ) $, but they don't need to know that.

If you don't like that, hardmath has given an answer; but you can make it look more like $ \sin ( 1 / x ) $ (including being continuous) by using $ 2 \{ 1 / x \} - 1 $ when $ [ 1 / x ] $ is odd and $ 1 - 2 \{ 1 / x \} $ when $ [ 1 / x ] $ is even. Your students might prefer a more explicit piecewise-defined representation: $$ \cases { 1 - 2 / x & for $ x > 1 $, \\ 2 / x - 3 & for $ 1 / 2 < x \leq 1 $, \\ 5 - 2 / x & for $ 1 / 3 < x \leq 1 / 2 $, \\ \vdots & etc. } $$ This is based on a piecewise-linear approximation of the sine function (or rather a cosine function with period $ 2 $). You can make it differentiable by using a piecewise-quadratic approximation instead, twice differentiable using a piecewise-cubic, etc. (But to make it infinitely differentiable, you're back at the sine function.)

Toby Bartels
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  • But to make it infinitely differentiable, you're back at the sine function. --- Probably you meant "real-analytic", and not "infinitely differentiable". – Dave L. Renfro Jul 20 '21 at 20:01
  • You could make make it infinitely differentiable without making it analytic yet, but I don't think that that would be any simpler, so you may as well make it analytic. – Toby Bartels Jul 20 '21 at 21:56
  • I thought you were making a veiled reference to the identity theorem. For example, even agreement with the values of $\sin(1/x)$ at relatively few positive values of $x$ (a sequence of such having a positive limit point would suffice) suffices to agree with $\sin(1/x)$ for all positive values of $x$ (if one assumes the function is analytic for all positive values of $x).$ – Dave L. Renfro Jul 20 '21 at 22:05
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    No, I wasn't trying to say that you'd be forced to use a sine function, only that there'd be nothing simpler to use by then. (Besides, we're only fixing the values at $1/n$, so the identity theorem doesn't apply.) – Toby Bartels Jul 21 '21 at 00:49
  • *so the identity theorem doesn't apply. --- Oops, I missed this! In fact, I missed the requirement that the domain of analyticity be connected when I was writing my comment and had to go back and change "nonzero values" to "positive values". I did remember from the beginning that the limit point needs to belong to the open set (otherwise this research would not exist), and I made sure to include that. However, after doing all this, I overlooked that your sequence's limit point does not belong to the open set . . . – Dave L. Renfro Jul 21 '21 at 16:09