How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

Question

How can one prove the statement $$\lim\limits_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.

This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated.

Answer 1

The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$. The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$. By inclusion, we get $$ \frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1} $$ Dividing $(1)$ by $\frac{1}{2}\sin(x)$ and taking reciprocals, we get $$ \cos(x)\le\frac{\sin(x)}{x}\le1\tag{2} $$ Since $\frac{\sin(x)}{x}$ and $\cos(x)$ are even functions, $(2)$ is valid for any non-zero $x$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Furthermore, since $\cos(x)$ is continuous near $0$ and $\cos(0) = 1$, we get that $$ \lim_{x\to0}\frac{\sin(x)}{x}=1\tag{3} $$ Also, dividing $(2)$ by $\cos(x)$, we get that $$ 1\le\frac{\tan(x)}{x}\le\sec(x)\tag{4} $$ Since $\sec(x)$ is continuous near $0$ and $\sec(0) = 1$, we get that $$ \lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5} $$

Answer 2

You should first prove that for $x > 0$ small that $\sin x < x < \tan x$. Then, dividing by $x$ you get $$ { \sin x \over x} < 1 $$ and rearranging $1 < {\tan x \over x} = {\sin x \over x \cos x }$ $$ \cos x < {\sin x \over x}. $$ Taking $x \rightarrow 0^+$ you apply the squeeze theorem. For $x < 0$ and small use that $\sin(-x) = -x$ so that ${\sin(-x) \over -x} = {\sin x \over x}$.

As far as why the first inequality I said is true, you can do this completely from triangles but I don't know how to draw the pictures here.

Answer 3

It depends on your definition of the sine function. I would suggest checking out the geometric proof in ProofWiki.

Answer 4

Look this link

http://fatosmatematicos.blogspot.com/2010/08/provas-sem-palavras-parte-20.html

Answer 5

Usually calculus textbooks do this using geometric arguments followed by squeezing.

Here's an Euler-esque way of looking at it---not a "proof" as that term is usually understood today, but still worth knowing about.

Let $\theta$ be the length of an arc along the circle of unit radius centered at $(0,0)$, from the point $(1,0)$ in a counterclockwise direction to some point $(\cos\theta,\sin\theta)$ on the circle. Then of course $\sin\theta$ is the height of the latter point above the $x$-axis. Now imagine what happens if $\theta$ is an infinitely small positive number. Then the arc is just an infinitely short vertical line, and the height of the endpoint above the $x$-axis is just the length of the arc. I.e. when $\theta$ is an infinitely small number, then $\sin\theta$ is the same as $\theta$. It follows that when $\theta$ is an infinitely small nonzero number, then $\dfrac{\sin\theta}{\theta}=1$.

That is how Euler viewed the matter. See his book on differential calculus.

Answer 6

Here you may see an elementary approach that starts from a very interesting result, see this problem. All you need is a bit of imagination. When we take $\lim_{n\rightarrow\infty} \frac{n\sin(\frac{\pi}{n})}{1+\sin(\frac{\pi}{n})}$ we may notice that we have infinitely many circles surrounding the unit circle with infinitely small diameters that lastly perfectly approximate the length of the unit circle when having it there infinity times . Therefore when multiplying n by the radius under the limit to infinity we get π. Let's denote $\frac{\pi}{n}$ by x.

$$\lim_{x\rightarrow0}\frac{\pi\sin(x)}{x(1+\sin(x))}={\pi}\Rightarrow\lim_{x\rightarrow0}\frac{\sin(x)}{x(1+\sin(x))}=1\Rightarrow\lim_{x\rightarrow0}\frac{\sin(x)}{x}=1$$

The proof is complete.

Answer 7

I think that all proofs above are beautiful but I cannot see why we have to complicate it. We know that for small values of x that sinx is approximately equal to x. If now let x in sinx go to zero it means that sinx goes to x. That is, our "error" or approximation goes to zero. We divide this by x and we have 1. Is not this valid?

Only thing is that this does not look like the typical proof.

Answer 8

I am not sure if it counts as proof, but I have seen this done by a High Schooler.

In the given picture above, $\displaystyle 2n \text{ EJ} = 2nR \sin\left( \frac{\pi}{n } \right ) = \text{ perimeter of polygon }$.

$\displaystyle \lim_{n\to \infty }2nR \sin\left( \frac{\pi}{n } \right ) = \lim_{n\to \infty } (\text{ perimeter of polygon }) = 2 \pi R \implies \lim_{n\to \infty}\frac{\sin\left( \frac{\pi}{n } \right )}{\left( \frac{\pi}{n } \right )} = 1$ and let $\frac{\pi}{n} = x$.