Why is lim sinx x 1

It depends on your definition of the sine function. I would suggest checking out the geometric proof in ProofWiki. This can easily be done using the picture below. Here is a different approach that uses the integral definition of the arcsine function. We will deduce the limit of interest without appeal to geometry or differential calculus. Instead, we only rely on elementary analysis of continuous functions and their inverses along with simple properties of the Riemann integral.

To that end, we now proceed. Since the circle sector is a subset of the quadrilateral, the perimeter of the circle sector is less than the perimeter of the quadrilateral. In particular. The convexity of the disk follows from the triangle inequality: a disk is a closed ball for the euclidean distance.

Similarly we write three systems for the initial conditions:. The proof is completed when the Wallis product is used. Simple one is using sandwich theorem Which demonstrated earlier.

You can use geogebra to see the visualization of this phenomena using geogebra. Here is a more fun one! This is a new post on an old saw because this is one of those things where that I can see how that, all too sadly, the way in which we've structured the current maths curriculum really doesn't make it possible to do these kinds of things the justice they deserve and I think, ultimately, that is a disservice to many learners.

The truth is, this limit cannot really be given an honest proof without an honest definition of the sine function, first. And that is not as easy as it seems. Even if we consider the simple notion from many trigonometric treatments that the sine is equal to the "length of the opposite side of the right triangle divided by the length of its hypotenuse", that doesn't truly solve the problem because there is actually a subtle missing element and that is that sine is not a function of a "right triangle" though you could define that if you wanted, and it'd be easy!

And actually parsing out what "angle measure" means, it turns out, is essentially equivalent to defining the sine function in the first place, so this approach is circular!

So how do we define sine, or angle measure? Unfortunately, any approach to this is such that it must involve calculus. Then we have the famous "impossible" problem of "trisection of the angle" which vexed even the ancient Greeks and for which people would keep trying to pound at until Pierre Wantzel finally proved it undoable over two thousand years after. We are asking for a mathematical widget that can not only trisect, but 5-sect, sect, etc.

Indeed, not only is the sine function not trivial, we could argue that even the exponential function is considerably easier to treat than sine, though I won't give such a treatment here. Thus, how do we do it? Well, the key observation is that our "steady" angle measure is one which is, effectively, defined by the arc length of a segment of circle intercepted by the angle when drawn at the circle's center and projected outward. In particular, this should be "obvious" from the circularly-introduced geometric formula.

Hence, we will begin with the arc question first and one will see that this answer will end up using a fair bit of Calculus II material to answer this Calculus I-level question about a supposedly pre-Calculus mathematical object. Indeed, this is what the whole "radian measure" is: it's a measure of angles in terms of the arc length of the piece they cut from a unit circle i.

Thus, what we have above is something called an arc length parameterization of the circle - and that tells us how we need to proceed. First, we need a separate definition of the arc length of a circle. How do we get that?

Well, we will obviously need a more elementary circle equation, first, than the one we just gave, and that means going to the simple algebraic definition,. And now this is where we then must introduce Calculus II-level concept - namely, integration for arc length. Since the "real", or base, function here is really the inverse function, i. By the power rule and chain rule,.

What I am saying is that, in fact, that is not truly honestly possible and reveals a weakness of the curriculum in that it doesn't actually follow the proper logical buildup of the mathematical edifice. What really should be done is to leave trig for later , that is, skip trig and go for Calculus first. When I studied maths on my own, I did just that. In fact, I'd say, as many educators have suggested, that most people don't need either, but really need more statistics instead.

Then for those who do pursue higher maths, if we've done algebra and statistics, we already have right there a lot of interesting material we can build on for calculus, including the exponential function. That said, rote crunching is not something I suggest banning either but I suggest that ideas, concepts, and creativity should come first, then you get into those techniques because very often they are also still useful in analysis and being fluent at them can also make you able to solve problems more quickly, e.

This is a variant of robjohn's answer. One possible definition is here. This allows us to take the limit inside and we get. This is not a rigorous proof, but is instead an intuitive argument. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Ask Question.

Asked 10 years ago. Active 12 days ago. Viewed k times. Show 11 more comments. Active Oldest Votes. Also, I didn't have a problem with using that arclength and area are proportional to the angle.

Show 24 more comments. Domates 4 4 silver badges 16 16 bronze badges. The definition of radians makes the picture above true. For whatever reason, this is the proof I like the most because it relates the tangent line at the point on the circle to the value we call "tangent". To each his own Show 4 more comments. That is how Euler viewed the matter.

See his book on differential calculus. Michael Hardy Michael Hardy 1. Was it very difficult to read because of the notation and language of the time? I wouldn't say the differences in language and notation were the challenging part. Show 1 more comment.

Anshul Laikar 6 6 bronze badges. The picture isn't immediately convincing. Add a comment. And so I can just write that down as the absolute value of the tangent of theta over two. Now, how would you compare the areas of this pink or this salmon-colored triangle which sits inside of this wedge and how do you compare that area of the wedge to the bigger triangle?

Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge and the area of the wedge is less than or equal to the area of the big, blue triangle. The wedge includes the salmon triangle plus this area right over here, and then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true and I'm just gonna do a little bit of algebraic manipulation.

Let me multiply everything by two so I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta which is less than or equal to the absolute value of tangent of theta, and let's see. Actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's gonna be the same thing as the absolute value of tangent of theta.

And the reason why I did that is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that I'm gonna divide this by an absolute value of sine of theta.

I'm gonna divide this by an absolute value of the sine of theta and then I'm gonna divide this by an absolute value of the sine of theta.

And what do I get? Well, over here, I get a one and on the right-hand side, I get a one over the absolute value of cosine theta. These two cancel out. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities.

The reciprocal of one is still going to be one but now, since I'm taking the reciprocal of this here, it's gonna be greater than or equal to the absolute value of the sine of theta over the absolute value of theta, and that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta.

We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there, so that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. And if we're in the fourth quadrant and theta's negative, well, sine of theta is gonna have the same sign. It's going to be negative as well.

And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value, so I can erase those.

If we're in the first or fourth quadrant, our X value is not negative, and so cosine of theta, which is the x-coordinate on our unit circle, is not going to be negative, and so we don't need the absolute value signs over there.

Now, we should pause a second because we're actually almost done. We have just set up three functions. You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that.

And over the interval that we care about, we could say for negative pi over two is less than theta is less than pi over two, but over this interval, this is true for any theta over which these functions are defined. Sine of theta over theta is defined over this interval, except where theta is equal to zero.

But since we're defined everywhere else, we can now find the limit. So what we can say is, well, by the squeeze theorem or by the sandwich theorem, if this is true over the interval, then we also know that the following is true.

And this, we deserve a little bit of a drum roll. The limit as theta approaches zero of this is going to be greater than or equal to the limit as theta approaches zero of this, which is the one that we care about, sine of theta over theta, which is going to be greater than or equal to the limit as theta approaches zero of this.

Now this is clearly going to be just equal to one. This is what we care about. And this, what's the limit as theta approaches zero of cosine of theta?

Well, cosine of zero is just one and it's a continuous function, so this is just gonna be one. So let's see. This limit is going to be less than or equal to one and it's gonna be greater than or equal to one, so this must be equal to one and we are done.

Squeeze theorem.