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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 1

Lesson 9: Determining limits using the squeeze theorem

# Limit of sin(x)/x as x approaches 0

Showing that the limit of sin(x)/x as x approaches 0 is equal to 1. If you find this fact confusing, you've reached the right place!

## Want to join the conversation?

• how did one come up with such a complex proof ?
i mean how did one guess that we must first take a part of circle cut it into triangles and make an inequality
also the whole proof was making no sense until we reached the last step ......

sorry if my question is stupid or if i am missing some basic skills to come up with a proof like this! •   whoever did this really clever theorem didn't made it by accident or just because he wanted to know whats the limit of sinx/x without any previous knowledge. It is enough to see the graph of the function to see that sinx/x could be 1. NOW, that's the first step. then, the mathematician must figure a formal and irrefutable theorem for the limit to be commonly accepted. It must be in fact hard to find some relationship like this but these people are just committed with their work but they are normal just like you.
• It seems clear that the area of the wedge is greater than the red triangle, but how do we know they can be equal? Same with the blue triangle being greater than or equal to the wedge. •  As θ approaches 0, all the three areas approach each other, try to think about it visually. Maybe that is the reason we put an equality sign there.
• why does Sal only consider the first and fourth quadrant? he doesn't do an restrict the values of theta. •   Sal was trying to prove that the limit of sin x/x as x approaches zero. To prove this, we'd need to consider values of x approaching 0 from both the positive and the negative side. So, for the sake of simplicity, he cares about the values of x approaching 0 in the interval (-pi/2, pi/2), which approach 0 from both the negative (-pi/2, 0) and the positive (0, pi/2) side. The interval (-pi/2, pi/2) spans only the first and fourth quadrants. This is why Sal discards the rest.
• To Sal Khan,

I believe that more important than learning a concept itself is learning how it came to be, learning how it was discovered, learning how to THINK like that. You can just teach the proof just because you have learnt the proof. But please, if possible, teach how to think like a mathematician. Teach how we can, by ourselves, prove something like this.
I mean, that proof made no sense in the beginning. Half the time, I didn't have a single clue to what he was doing or more importantly, 'why' he was doing it.

Nobody's a genius mathematician by birth (well except for some maybe). You can't just think of proofs like that out of thin air! Please tell me, I want to know, of some way to actually think of doing bizarre things like that to prove what we want. I think that remembering the proof after someone else has taught me is kinda useless. •   Math major here.

Your objection is well-placed, and you're correct that proof-writing is a skill that is mostly improved by practice. Before you continue, there are two unfortunate truths to keep in mind:
1. The United States math curriculum places almost zero emphasis on proof writing or proof comprehension (and the things that pass for proofs in geometry are a joke). Sal's primary target audience is U.S. grade-schoolers and high-schoolers, and so the videos on Khan Academy cater to these admittedly bad standards. Before Common Core became widespread, it was common to see KA videos presenting the exact same topics in the exact same order as your own teacher.

2. The culture in the mathematics community dictates that once you've written a proof, you 'polish' it to make it as short and concise as possible. The onus is very much put on the reader of the proof to slog through it word-by-word, and not on the writer to be clear.

All that said, the only thing that Sal really pulled out of a hat was the idea to compare the different areas in the figure. To draw the figure in the first place was fairly natural, because how else can we interpret the sine function if not by the unit circle? And once he had made the decision to compare areas, the proof was fairly straightforward algebra.

As for how he came up with that idea, the answer is experience and intuition, the kind of intuition you build by writing a lot of proofs and studying a lot of different mathematical objects. If you really want to learn this type of thinking, a standard place to start is the book 'How to Solve It' by George Pólya.
• Why were the inequalities switched? •   Because he took the reciprocal of each function/number. The same way the reciprocal of three is 1/3, the reciprocals of each function (other than one) were less than the original value. Thus, the inequalities needed to be switched. Think of how 1 is less than three, which is less than four, but if you take the reciprocals you switch the order of inequalities. (1 is greater than 1/3, which is greater than 1/4). Hope this helps!
• Why were the absolute value signs necessary for inclusion in the beginning and middle parts of the process but discarded at the end? At he mentions that they are discarded because we're only concerned with the 1st and 4th quadrants but why is that? Please explain. • Sal uses absolute values because we're dealing with areas and areas must be positive. In the instance that θ is negative, we'd have negative area since sine and tangent of θ would be negative. So, he took the absolute values of sine and tangent so that the areas would be positive regardless of θ.

Now, why did he discard the absolute values?

1. In the case of |sin(θ)|/|θ|
Well,
sin(θ)/θ = sin(-θ)/-θ = -sin(θ)/-θ

So, |sin(θ)|/|θ| = sin(θ)/θ

He can therefore discard the absolute values.

2. In the case of cos(θ)

Since -π/2 < θ < π/2,

cos(θ) is always positive and so for this constraint, |cos(θ)| = cos(θ).

So the absolute values can be discarded
• At why is the limit of 1 is 1 when theta approaches zero? • The lim(1) when Θ→0 means: on the graph y=1, what does the y-coordinate approach when the x-coordinate (or in this case Θ) approach 0. But on the graph y=1, the y-coordinate is always 1 no matter what the x-coordinate is. In other words, lim(k) as Θ→n = k, where k,n are any real numbers.   