# sin(1/x)

In my most recent calculus classes, I wanted to show my kids their first “not nice” functions. After being introduced to how to find limits graphically (fancy way of saying: looking at the graph of a function) and numerically (fancy way of saying: using the graphing calculator’s TABLE function to guesstimate limits), I wanted to have them think about what they learned.

I had time to show one class that these methods aren’t foolproof — that the calculator can lie to you, and make you think a limit is 3 when it is in fact 3.004, or that it can’t graph things when numbers get too large or too small. So they have to be careful. And that we will be learning algebraic methods to do limits. But for now, they need to use their brains and wits.

So I divided them into groups of 2 and 3 and had them use whatever methods they wanted to find:

$\lim_{x \to 0} \sin(\frac{1}{x})$

I made them each draw a sketch of the function, write down an appropriate table of values, make observations about the function, and then decide on an answer. (In one class, I had each group turn in their findings, and then I photocopied them and distributed them and had the class talk collectively about the results the next day. In the other class, we didn’t have time for this, and we just met up together as a group to talk.)

FYI, the graph is here.

It was great. Students were debating whether the craziness was a function of the calculator lying or if that actually was what the function looked like. They wondered if the limit was 0 or if it was “does not exist.” They noticed that the function starts to oscillate more and more rapidly as $x$ approaches 0. They noticed that it bounced between -1 and 1. It’s not an easy question to solve with this information.

When we came back as a group, we talked about their observations and conclusions, and documented them on the board — so everyone had the same notes. Then I said: “so… one of you said that the function is crossing the $x$ axis more and more as $x$ is getting closer and closer to 0. Can we be more exact? Where does the function cross the $x$ axis?”

Of course my students didn’t know exactly what to do. We got to the point where we knew we had to solve:

$\sin(\frac{1}{x})=0$

But then they were stuck. So I guided them through it.

I asked: “When is $\sin(\square)=0$

We generated: $\square=...,-4\pi,-3\pi,-2\pi,-1\pi,0,1\pi,2\pi,3\pi,4\pi,...$

We then said: $\frac{1}{x}=...,-4\pi,-3\pi,-2\pi,-1\pi,0,1\pi,2\pi,3\pi,4\pi,...$

We went through solving one of the equations for $x$ and saw that we needed the reciprocals…

We concluded: $x=...,\frac{-1}{4\pi},\frac{-1}{3\pi},\frac{-1}{2\pi},\frac{-1}{1\pi},\frac{1}{\pi},\frac{1}{2\pi},\frac{1}{3\pi},\frac{1}{4\pi},...$

I then asked: So what? Why did we do this? Don’t lose the forest for the trees…

Finally, we converted those numbers to decimal approximations

$x \approx \pm 0.318,\pm 0.159,\pm 0.106, \pm 0.080, \pm 0.064, \pm 0.053, \pm 0.045, \pm 0.040, ...$

and saw that the zeros were getting more and more frequent as we approached 0. No matter how close we came to zero, we were still going to be bobbing up and down on the function. And crucially, we’ll be bobbing up and down between $-1$ to $1$.

We then talked about what a limit means again… what the $y$ value of a function is approaching as the $x$ value gets closer and closer to a number. Using that informal definition, I asked them if the $y$ value of the function was approaching some number as $x$ was approaching 0.

At this point, most of my kids had that “a hah” moment.

I am definitely doing this again next year, but perhaps more formalized. I might generate a list of good conceptual questions to walk them through this more systematically. One such question: “How many zeros are there in the interval (.5,1)? How about (.1,1)? How about (.01,1)? How about (.001,.1)? How about (0.0001,1)? And finally, how about (0,1)?” Another such question: “How do we know the function will bounce between $-1$ and $1$?”

Also, maybe next year, I’ll couple it with an analysis of the function:

$\lim_{x \to 0} \sin(x)\cos(\frac{1}{x})$

The function behaves similarly (crosses the $x$ axis more and more rapidly as $x$ approaches 0), but the limit in this case is 0. You can see it in the graph easiest.

So if anyone out there is looking for something to spice limits up, you might want to really go in depth into these functions. They are often used as exemplars, but rarely investigated.

1. I feel like I’ve been left hanging. It’s been a long time since I calculated limits. Is the answer that, in fact, there is no limit because the value will bounce more wildly between -1 and 1 as x approaches 0? I also wonder if the guided questions will narrow their thinking and make the activity more of a worksheet or will generate more a-ha moments because of the more formal pencil-and-paper nature instead of the social aspect of your questioning. Thanks for sharing.

2. @Nick: Sorry, the answer is the limit does not exist.

Yeah, I was thinking about a guided sheet too for next year (doing something more formalized). But there was something awesome about the fact that they were given something out of the blue, and they have to just sit back and come up with observations. I’ll probably start out th esame way, and when we make the two big observations, we’ll start digging.

3. @Nick: Sorry, the answer is the limit does not exist.

Yeah, I was thinking about a guided sheet too for next year (doing something more formalized). But there was something awesome about the fact that they were given something out of the blue, and they have to just sit back and come up with observations. I’ll probably start out the same way, and when we make the two big observations, we’ll start digging with a worksheet.

4. Hi Sam – would it be OK with you if I include this post in next week’s math teachers at play blog carnival?
Thanks –
— Dan

5. Thanks Sam. I was thinking the same thing, that the genuine interest you show in questioning and working with their ideas is possibly more engaging and satisfying then the worksheet. Thanks for the help.

6. Eric Jablow says:

When you cover differentiation, you should show the corresponding example, $x^{4} \sin {\frac{1}{x^{3}}}$. Ask your students where the function is differentiable, and then ask the students where the derivative is continuous.

This is why we’re careful to define functions of class $C^{n}$ as functions with $n$ continuous derivatives. Of course, all but the last are obvious. Then, you can point out that if a function is differentiable everywhere, its derivative need not be continuous, but it is continuous on a dense subset of the real numbers. Its points of discontinuity are of first Baire category, even. Okay, maybe not that for high school students. But you can point out that even though it need not be continuous, it does satisfy the intermediate value theorem.

7. Jessica J. says:

So I know that the answer is that the limit does not exist, but I am still confused as to WHY. Could you help me out with that?

1. The limit — informally — is the single number (y-value) that a function is approaching as the x-value is getting closer and closer to 0. But as x gets closer and closer to 0, the y value [sin(1/x)] bounces between 1 and -1 more and more frequently. The function does not settle down to a single y value, but starts oscillating wildly as x gets closer and closer to 0. Since the function is not approaching a single y-value as x approaches 0, the function does not have a limit.

8. Mohini says:

So I can see I’m 4 months late in my handclapping- but this is awesome!!!
I’m a newly inducted undergrad in maths and my first few days of college left me feeling like a lobotomy would be preferable.
And you explained this graph so beautifully that I remembered why I took up math in the first place. Your students are lucky.
So…thank you.

1. Wow, those are kind words! Thank you for taking the time to leave them. I know my students were a little frustrated with the uncertainty and weirdness of the function, so they might not think they were lucky to go through this exercise — but I know they learned from it.

Sam

9. Kim says:

I am studying to be a math teacher – and my husband works at an International IB school – this is a fantastic example of inquiry based learning – heaps of higher order thinking! It’s a real challenge to do this – so thanks for some brilliant ideas! Kim

10. this is absolutely amazing tutoring, what pure mathematics is all about, observation and inference. Absolutely what i would try and follow up!@

11. heifer says:

This is my first year teaching AP Calculus…after teaching grades 9-12, algebra to pre-cal for 19 years, I finally made the jump. My concern is not what to teach, but *HOW* and you have some amazing insights. Beg, borrow, steal…and you have given me a good boost of confidence that I can present an interesting and insightful lesson. Thanks!

1. That’s super kind of you! I love teaching calculus… but I am a pale shadow of Bowman Dickson (www.bowmandickson.com) so you should check him out. Also, because he teaches AP calculus AND non-AP calculus, so he’ll have insights on the AP curriculum that I don’t even get to.

And yes: beg, borrow, steal. BEST THING EVAR!

12. great prompt. and I agree the roots are the easiest way to grok it.

13. I love this post! I am a preservice math teacher from Michigan, and I am learning about ways to get students to discover knowledge and get them thinking conceptually. This is such an incredible way to introduce limits! I want to teach calc now just so I can do this exact same lesson :)

Was this the first time that students had been introduced to a limit that does not exist?

1. Congrats on working to becoming a teacher! Huzzah!

To answer your question: no, students had learned about simple limits that didn’t exist — at vertical asymptotes or jump discontinuities. But this was the first weird example that wasn’t one of those basic ones!

14. Larry Wu says:

Can’t you find the limit by using the Squeeze Theorem?

Since -1 ≤ sin(1/x) ≤ 1, -x ≤ x sin(1/x) ≤ x, for positive values of x. Then, since both x and -x have limits of 0 when x approaches 0, the limit of sin(1/x) as x approaches 0 must be 0.

I am not sure, maybe I did something wrong.

1. This doesn’t work, I don’t think. All you’ve shown is that as x->0, that you know x*sin(1/x) goes to 0… not that sin(1/x) goes to 0.

2. J says:

Consider $1/x$ and $x^2$: $x^2 * 1/x \rightarrow 0$ as $x \rightarrow 0$, but $1/x \not\rightarrow 0$.

15. manas says:

When we’re talking about lim(h->0) sin1/h ……it will be a really big number…..so you should have told about how graph looks there