# Prove the inequality $\frac{\sin(x)}{x}<1$

Is there a good way to show that $\frac{\sin(x)}{x}$ is bounded above by $1$?

We can see visually that $\frac{\sin(x)}{x}$ is bounded above by $1$ because the tallest hump is at the origin and $\lim_{x \to 0} \frac{\sin(x)}{x}=1$. But is there a way to prove this rigorously? Preferably without expanding $\sin(x)$ into a Taylor series, unless Taylor series is the only way.

• You just need for $x>0 \implies \sin x < x$. Commented Apr 3, 2016 at 4:56

Clearly, $|\sin(x)|\leq |x|$ when $|x|>1$. For $|x|\leq 1$, you can use the mean value theorem to show that $x=0$ is the only solution to $\sin x=x$. Thus, $\sin x<x$ for $x>0$ and $\sin(x)>x$ for $x<0$. This implies that $$\left|\frac{\sin(x)}{x}\right|\leq 1.$$

• Like in the other deleted post, you just assumed what you are trying to prove, I think, by saying $|\sin(x)| \le |x|$ because that is synonymous to saying $|\frac{\sin(x)}{x}| \le 1$... Commented Apr 3, 2016 at 5:37
• I perhaps. I did skip the detail that $|\sin (x)|\leq1$, but once this is known, it is obvious that the inequality holds for $|x|\geq 1$. Do you feel that $|\sin (x)|\leq 1$ needs to be proven? Commented Apr 3, 2016 at 5:41
• Oh, I see, you wrote $|\sin(x)| \le 1 < |x|$ for $|x| > 1$, and the result follows for $|x| > 1$. Now I just have to address the $|x| \le 1$ case using the MVT. Commented Apr 3, 2016 at 5:46

Consider this image of the unit circle where the area of the triangle is $$\frac{1}{2}\sin(\theta)$$ and the area of the circle sector is $$\frac{1}{2}\theta$$.

You can see that $$\frac{1}{2}\sin(\theta) \le \frac{1}{2}\theta.$$

• +1 Absolutely brilliant! One of the best visual proofs I've ever seen! (Posted a whole 7 years later lol) Commented Dec 15, 2023 at 15:23

Use $\sin'(x) =\cos(x)$, and $\cos(x) \le 1$ to get, for $x \ge 0$, $\sin(x) =\int_0^x \cos(t)dt \le \int_0^x 1dt =x$.

L'Hopital's rule works: $$\lim_{x \to 0}\frac{\sin(x)}{\ x}$$ Becomes $$\lim_{x \to 0}\frac{\cos(x)}{\ 1}$$ Evaluating to $$1$$

It may not be as rigorous as you want though.

• l'Hopital doesn't imply boundedness by $1$. Commented Apr 3, 2016 at 5:07