WEBVTT
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Investigate the behavior of π of π₯ equals two cos one over π₯ as π₯ tends to zero.
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The first part of this question is to complete the table of values of π of π₯ for values of π₯ that get closer to zero.
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And if we look at the values of π₯ that weβre given in the table, we can see that they really do get closer and closer to zero.
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One over 99π is already quite small.
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But one over 100π is even smaller, even closer to zero.
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And the trend continues, one over 999π is smaller still.
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Okay, now letβs go about completing this table of values.
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In the first cell we have to fill, we need to put the value of π of one over 99π.
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But, what is this value?
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We find it by using the definition of the function π.
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π of π₯ is two cos one over π₯.
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And so, π of one over 99π is two cos one over one over 99π.
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One over one over 99π is just 99π.
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And so, we can simplify to get two cos 99π.
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Whatβs the value of cos of 99π?
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The function cos is a periodic function.
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And its period β assuming weβre working in radians, which we are β is two π.
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Another way of saying this is that the value of cos of π₯ plus two π is the same as the value of cos of π₯.
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And by applying this rule twice, we can see that cos π₯ plus four π is equal to cos π₯ plus two π which is equal to cos π₯.
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By repeating this process, we can see that cos π₯ plus two ππ, for any natural number π, is equal to cos π₯.
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And in fact, by working backwards, we can show that this holds for negative integers π too.
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And so, the result holds for any integer π.
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Okay, but how does this result help us simplify two cos 99π?
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Well, we can write 99π as π plus two π times 49.
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And so, using our rule, we get two cos π.
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Another way of thinking about this is that if you keep subtracting two π from 99π, eventually, youβll end up with just π.
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As cos π is just negative one, two times cos π is negative two.
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And we found the value of π of one over 99π.
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Now, we need to find the next value in our table.
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The process is almost exactly the same.
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We use the definition of π and the fact that one over one over 100π is 100π.
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But this time, as 100π is an even multiple of π, we eventually get down to zero when we keep subtracting two π from it.
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We therefore find that π of one over 100π is two cos zero.
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And as cos zero is one, two cos zero is two.
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We put this two into our table and move on to the next entry.
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π of one over 999π is two cos one over one over 999π which is two cos 999π.
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And as 999 is an odd number, the value of cos 999π is the value of cos π.
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And just like with π of one over 99π, as cos of π is negative one, π of one over 999π is negative two.
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Rather than using the same procedure to fill in the remaining three gaps in our table, weβre going to use the pattern that weβve discovered.
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When π₯ is a unit fraction with an odd multiple of π in its denominator, then the value of π of π₯ is negative two.
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We can prove this if weβd like to.
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Any odd number can be expressed as two π plus one, where π is some natural number.
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And following our procedure, we can see that π of one over two π plus one π is negative two.
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Pause the video if youβd like to think about this some more.
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And so, in particular, as 9999 is an odd number, when π₯ is one over 9999π, then π of π₯ is negative two.
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So that covers the case when the denominator of our unit fraction is an odd multiple of π.
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But what about the even multiples of π?
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Writing our general even number as two times π, we see that when π₯ is one over an even multiple of π, π of π₯ is two.
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And so, both π of one over 1000π and π of one over 10000π are both two.
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We might also like to summarise the discoveries weβve made, that π of one over two ππ is two and π of one over two π plus one π is negative two, for any natural number π.
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Letβs now move on to the next part of the question.
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What does this suggest about the graph of π close to zero?
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Well, weβve seen for some very small values of π₯ in our table that π of π₯ can be negative two or two.
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And weβve seen from our discoveries that π of one over two ππ is two and π of one over two π plus one π is negative two.
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That however close you are to zero, you can find a smaller value of π₯ for which π of π₯ is two or π of π₯ is negative two.
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So what does this suggest about the graph of π close to zero?
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It suggests that it oscillates between negative two and two.
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You might like to just check that this makes sense, given the definition of our function π, that π of π₯ equals two cos one over π₯.
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Whatever the value of π₯ is, one over π₯ is just some number.
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And so, π of π₯ is two times cos something which is bounded between negative two and two, inclusive.
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So certainly, our function canβt be oscillating between two values greater in absolute value.
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Also, looking at the table you might be tempted to think that π of π₯ can only take two possible values, negative two or two.
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But actually, the function takes values between negative two and two as well.
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Itβs just that the values of π₯ in our table were chosen to produce these extreme values.
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You might like to think about how to find values of π₯ for which π of π₯ is zero or one or negative 1.5.
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Moving on to the third and final part of the question, hence, evaluate the limit of π of π₯ as π₯ approaches zero.
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Well, weβve seen that as π₯ approaches or tends to zero, π of π₯ oscillates between negative two and two.
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The value of π of π₯, therefore, is not getting closer and closer to some real number.
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And so, the limit of π of π₯ as π₯ approaches zero cannot be any real number.
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And as π is a bounded function, taking values between negative two and two, the limits cannot be infinite either.
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The only conclusion we can draw is that the limit does not exist.
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Now, the purpose of this question was to work all this out without relying on a graph.
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But now that we have worked it out, you might like to use a graphing calculator or graphing software to explore the graph of π.
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As you might be able to see from the small graph Iβm showing on screen, something pretty wild happens near the origin.