Video: Limits and Oscillating Behavior

Investigate the behavior of 𝑓(π‘₯) = 2 cos (1/π‘₯) as π‘₯ tends to 0. Complete the table of values of 𝑓(π‘₯) for values of π‘₯ that get closer to 0. What does this suggest about the graph of 𝑓 close to zero? Hence, evaluate lim_(π‘₯ β†’ 0) 𝑓(π‘₯).

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Video Transcript

Investigate the behavior of 𝑓 of π‘₯ equals two cos one over π‘₯ as π‘₯ tends to zero.

The first part of this question is to complete the table of values of 𝑓 of π‘₯ for values of π‘₯ that get closer to zero. And if we look at the values of π‘₯ that we’re given in the table, we can see that they really do get closer and closer to zero. One over 99πœ‹ is already quite small. But one over 100πœ‹ is even smaller, even closer to zero. And the trend continues, one over 999πœ‹ is smaller still. Okay, now let’s go about completing this table of values.

In the first cell we have to fill, we need to put the value of 𝑓 of one over 99πœ‹. But, what is this value? We find it by using the definition of the function 𝑓. 𝑓 of π‘₯ is two cos one over π‘₯. And so, 𝑓 of one over 99πœ‹ is two cos one over one over 99πœ‹. One over one over 99πœ‹ is just 99πœ‹. And so, we can simplify to get two cos 99πœ‹. What’s the value of cos of 99πœ‹? The function cos is a periodic function. And its period β€” assuming we’re working in radians, which we are β€” is two πœ‹. Another way of saying this is that the value of cos of π‘₯ plus two πœ‹ is the same as the value of cos of π‘₯. And by applying this rule twice, we can see that cos π‘₯ plus four πœ‹ is equal to cos π‘₯ plus two πœ‹ which is equal to cos π‘₯.

By repeating this process, we can see that cos π‘₯ plus two πœ‹π‘›, for any natural number 𝑛, is equal to cos π‘₯. And in fact, by working backwards, we can show that this holds for negative integers 𝑛 too. And so, the result holds for any integer 𝑛. Okay, but how does this result help us simplify two cos 99πœ‹? Well, we can write 99πœ‹ as πœ‹ plus two πœ‹ times 49. And so, using our rule, we get two cos πœ‹. Another way of thinking about this is that if you keep subtracting two πœ‹ from 99πœ‹, eventually, you’ll end up with just πœ‹. As cos πœ‹ is just negative one, two times cos πœ‹ is negative two. And we found the value of 𝑓 of one over 99πœ‹.

Now, we need to find the next value in our table. The process is almost exactly the same. We use the definition of 𝑓 and the fact that one over one over 100πœ‹ is 100πœ‹. But this time, as 100πœ‹ is an even multiple of πœ‹, we eventually get down to zero when we keep subtracting two πœ‹ from it. We therefore find that 𝑓 of one over 100πœ‹ is two cos zero. And as cos zero is one, two cos zero is two. We put this two into our table and move on to the next entry.

𝑓 of one over 999πœ‹ is two cos one over one over 999πœ‹ which is two cos 999πœ‹. And as 999 is an odd number, the value of cos 999πœ‹ is the value of cos πœ‹. And just like with 𝑓 of one over 99πœ‹, as cos of πœ‹ is negative one, 𝑓 of one over 999πœ‹ is negative two. Rather than using the same procedure to fill in the remaining three gaps in our table, we’re going to use the pattern that we’ve discovered.

When π‘₯ is a unit fraction with an odd multiple of πœ‹ in its denominator, then the value of 𝑓 of π‘₯ is negative two. We can prove this if we’d like to. Any odd number can be expressed as two π‘˜ plus one, where π‘˜ is some natural number. And following our procedure, we can see that 𝑓 of one over two π‘˜ plus one πœ‹ is negative two. Pause the video if you’d like to think about this some more.

And so, in particular, as 9999 is an odd number, when π‘₯ is one over 9999πœ‹, then 𝑓 of π‘₯ is negative two.

So that covers the case when the denominator of our unit fraction is an odd multiple of πœ‹. But what about the even multiples of πœ‹? Writing our general even number as two times π‘˜, we see that when π‘₯ is one over an even multiple of πœ‹, 𝑓 of π‘₯ is two. And so, both 𝑓 of one over 1000πœ‹ and 𝑓 of one over 10000πœ‹ are both two. We might also like to summarise the discoveries we’ve made, that 𝑓 of one over two π‘˜πœ‹ is two and 𝑓 of one over two π‘˜ plus one πœ‹ is negative two, for any natural number π‘˜.

Let’s now move on to the next part of the question. What does this suggest about the graph of 𝑓 close to zero? Well, we’ve seen for some very small values of π‘₯ in our table that 𝑓 of π‘₯ can be negative two or two. And we’ve seen from our discoveries that 𝑓 of one over two π‘˜πœ‹ is two and 𝑓 of one over two π‘˜ plus one πœ‹ is negative two. That however close you are to zero, you can find a smaller value of π‘₯ for which 𝑓 of π‘₯ is two or 𝑓 of π‘₯ is negative two. So what does this suggest about the graph of 𝑓 close to zero? It suggests that it oscillates between negative two and two.

You might like to just check that this makes sense, given the definition of our function 𝑓, that 𝑓 of π‘₯ equals two cos one over π‘₯. Whatever the value of π‘₯ is, one over π‘₯ is just some number. And so, 𝑓 of π‘₯ is two times cos something which is bounded between negative two and two, inclusive. So certainly, our function can’t be oscillating between two values greater in absolute value. Also, looking at the table you might be tempted to think that 𝑓 of π‘₯ can only take two possible values, negative two or two. But actually, the function takes values between negative two and two as well. It’s just that the values of π‘₯ in our table were chosen to produce these extreme values. You might like to think about how to find values of π‘₯ for which 𝑓 of π‘₯ is zero or one or negative 1.5.

Moving on to the third and final part of the question, hence, evaluate the limit of 𝑓 of π‘₯ as π‘₯ approaches zero. Well, we’ve seen that as π‘₯ approaches or tends to zero, 𝑓 of π‘₯ oscillates between negative two and two. The value of 𝑓 of π‘₯, therefore, is not getting closer and closer to some real number. And so, the limit of 𝑓 of π‘₯ as π‘₯ approaches zero cannot be any real number. And as 𝑓 is a bounded function, taking values between negative two and two, the limits cannot be infinite either. The only conclusion we can draw is that the limit does not exist.

Now, the purpose of this question was to work all this out without relying on a graph. But now that we have worked it out, you might like to use a graphing calculator or graphing software to explore the graph of 𝑓. As you might be able to see from the small graph I’m showing on screen, something pretty wild happens near the origin.

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