# Why does $\lim_{x\rightarrow 0}\frac{\sin(x)}x=1$? [duplicate]

I am learning about the derivative function of $\frac{d}{dx}[\sin(x)] = \cos(x)$.

The proof stated: From $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$...

I realized I don't know why, so I wanted to learn why part is true first before moving on. But unfortunately I don't have the complete note for this proof.

1. It started with a unit circle, and then drew a triangle at $(1, \tan(\theta))$
2. It show the area of the big triangle is $\frac{\tan\theta}{2}$
3. It show the area is greater than the sector, which is $\frac{\theta}{2}$ Here is my question, how does this "section" of the circle equal to $\frac{\theta}{2}$? (It looks like a pizza slice).
4. From there, it stated the area of the smaller triangle is $\frac{\sin(\theta)}{2}$. I understand this part. Since the area of the triangle is $\frac{1}{2}(\text{base} \times \text{height})$.

5. Then they multiply each expression by $\frac{2}{\sin(\theta){}}$ to get $\frac{1}{\cos(\theta)} \ge \frac{\theta}{\sin(\theta)} \ge 1$

And the incomplete notes ended here, I am not sure how the teacher go the conclusion $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$. I thought it might be something to do with reversing the inequality... Is the answer obvious from this point? And how does step #3 calculation works?

## marked as duplicate by leonbloy, Ian, Milo Brandt, Mark Bennet, Did calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 18 '15 at 19:04

• Reversing the inequality is standard, but it really makes no difference. Do you know the squeeze theorem? – Vincenzo Oliva Jan 18 '15 at 18:25
• draw the three graphs $y = 1, y = 1/\cos x, y = x/\sin x$ and see how they behave around $x = 0.$ – abel Jan 18 '15 at 18:29

Draw the circle of radius $1$ centered at $(0,0)$ in the Cartesian plane.

Let $\theta$ be the length of the arc from $(1,0)$ to a point on the circle. The radian measure of the corresponding angle is $\theta$ and the height of the endpoint of the arc above the coordinate axis is $\sin\theta$.

Now look at what happens when $\theta$ is infinitesimally small. The length of the arc is $\theta$ and the height is also $\theta$, since that infinitely small part of the circle looks like a vertical line (you're looking at the neighborhood of $(1,0)$ under a microscope).

Since $\theta$ and $\sin\theta$ are the same when $\theta$ is infinitesimally small, it follows that $\dfrac{\sin\theta}\theta=1$ when $\theta$ is infinitesimally small.

That is how Leonhard Euler viewed the matter in the 18th century.

Why does the sector of the circle have area $\theta/2$?

The whole circle has area $\pi r^2=\pi 1^2 = \pi$. The fraction of the circle in the sector is $$\frac{\text{arc}}{\text{circumference}} = \frac{\theta}{2\pi}.$$ So the area is $$\frac \theta {2\pi}\cdot \pi = \frac\theta2.$$

• I love Euler, +1. Was he the first to conclude this (from a pre-calculus point of view)? – Vincenzo Oliva Jan 18 '15 at 18:35
• @VincenzoOliva : I doubt that Euler was the first to conclude this and I hesitate to call this a pre-calculus point of view. This is well within what today we consider calculus. To call it a precalculus point of view is to project our modern way of organizing the curriculum onto 18th-century thought. ${}\qquad{}$ – Michael Hardy Jan 18 '15 at 18:38
• I see, thank you. I said "precalculus" only because he did not have the modern notion of limit, for example he used $\infty$ in many of his calculations. Of course, the fact remains he probably was the mathematician who did more calculus than anyone else. And I do know the geometric proof of this notable limit being taught in high school nowadays is identical, except for the use of limits at the end (at least, my professor did use them). – Vincenzo Oliva Jan 18 '15 at 18:47
• @VincenzoOliva : If they're proving something about a limit, I wouldn't call it "pre-calculus". – Michael Hardy Jan 18 '15 at 18:51
• I agree that what he was doing was certainly not pre-calculus. I called "pre-calculus" his point of view. Do you think it was a calculus one? – Vincenzo Oliva Jan 18 '15 at 18:55

Do you knoe Squeez Theorem ?

You get $\frac{1}{\cos(\theta)} \ge \frac{\theta}{\sin(\theta)} \ge 1$. It is equivalent to $\cos(\theta) \le \frac{\sin(\theta)}{(\theta)} \le 1$.

Now, $\lim_{x \to 0} (\cos \theta) = 1$ and $\lim_{x \to 0} 1 = 1$. so, $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$. Here, it is an application of squeez theorem, so you must familiar with it.

• And why is #3 comes out to $\frac{\theta}{2}$? – George Jan 18 '15 at 18:31
• The inequalities reverse when you take the reciprocals. – Ian Jan 18 '15 at 18:31
• you have the same mistake like i – Dr. Sonnhard Graubner Jan 18 '15 at 18:32
• perhaps it is because we all do copy and paste... – Hardey Pandya Jan 18 '15 at 18:33

hint use that $$\cos(\theta)\le \frac{\sin(\theta)}{\theta}\le 1$$ and take the limit for $\theta$ goes to zero

• $\sin(\theta)/\theta$ is not greater than or equal to $1$, even in a neighborhood of $\theta=0$. (Everything works with the inequalities in the opposite direction, however.) – Ian Jan 18 '15 at 18:28
• I would say for the hint: "show that $\cos \theta\leq \frac{\sin \theta}{\theta}\leq 1$" – idm Jan 18 '15 at 18:33

Here is my question, how does this "section" of the circle equal to $\dfrac\theta2$ ? $($It looks like a pizza slice$)$

Well, to an arc length of $2\pi$ corresponds an area of $\pi=\dfrac{2\pi}2$ $($the full circle$)$. To an arc length of $\pi$ corresponds an area of $\dfrac\pi2$ $($the half-circle$)$. To an arc length of $\dfrac\pi2$ corresponds an area of $\dfrac\pi4$ $($the quarter-circle$)$. Etc. In general, to an arc length of $\alpha$ corresponds an area of $\dfrac\alpha2$. This follows from the symmetry of the geometric shape.