How to prove $|\sin x|\le |x|$ from $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$?

Question

Today one of my friend asked the following problem in class,

Suppose that we know $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$. Then can we prove geometrically that $|\sin x|\le |x|$ for all sufficiently small $x$?

The professor told that it seems unlikely that such a proof exists because the geometric proof of the limit $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$ uses the fact that $\cos x\le \dfrac{\sin x}{x}\le 1$ at some point.

After reading this post I have found that there are other proofs of the limit which doesn't use the inequality just mentioned. I tried to prove the claim but couldn't. Is there really any proof of the problem at all?

For the definition of $\sin$-function assume the definition given in the section Right-angled triangle definition of this page.

$\lim_{x\to 0} \sin(x) / x = 1$ alone won't imply $|\sin(x)| \le |x|$ for sufficiently small $|x|$. — user251257, Sep 29 '15 at 13:26
@user37238: Already said in the post. Please read it fully before commenting. — , Sep 29 '15 at 13:31
You also have $\lim_{x\to 0}\frac{x}{x+x^3}=1$ and you don't have $|x|\geq |x+x^3|$. In fact, $|x+x^3|=|x||1+x^2|\geq|x|$. — yurnero, Sep 29 '15 at 14:40
@yurnero: But in that case you don't have the properties of $\sin$. — , Sep 30 '15 at 15:33

score 2 · Accepted Answer · edited Sep 29 '15 at 14:36

2

Draw a unit circle. Draw a straight segment from the origin to a point $P$ on the edge of the circle at angle $x$ above the horizontal axis. Add a segment from $P$ straight down to the horizontal axis. The length of that segment is $|\sin(x)|$. The length of the arc from $P$ to $(1,0)$ is $|x|$. The arc is longer because the segment is the shortest possible path from P to the horizontal axis. QED, $|x|\ge |\sin(x)|$

edited Sep 29 '15 at 14:36

answered Sep 29 '15 at 14:31

Jerry Guern

2,662

"The length of the arc from $P$ to $(1,0)$ is $|x|$"-why? – Sep 29 '15 at 14:35
@user170039 If x is the angle in radians, then by definition of radians. Radians = archlength / radius – Jerry Guern Sep 29 '15 at 14:37
@user170039 That's why I specified "the unit circle", because the radius is conveniently 1. – Jerry Guern Sep 29 '15 at 14:38
@user170039 How did I deduce which relation? I didn't deduce the definition of radians, it's the definition. If the angle is x radians, that means that the arclength / radius is x. – Jerry Guern Sep 29 '15 at 14:40

score 1 · Answer 2 · answered Sep 29 '15 at 14:20

For a geometric proof of $|\sin x|\leq |x|$ you don't even have to know that limit. Consider an arc of length $2x$ on the unit circle. If its endpoints are $A$ and $B$ then certainly $|AB|\leq 2x$. On the other hand, by the "right-angled triangle-definition" of $\sin$ one has $|AB|=2\sin x$.

How to prove $|\sin x|\le |x|$ from $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$?

2 Answers2

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