# Function $\frac{\sin x}{x}$ [duplicate]

is the function $$f:\mathbb{R}\to\mathbb{R}$$ defined by $$f(x)=\frac{\sin x}{x}$$ for $$x\neq0$$ and $$f(0)=1$$ bounded on $$\mathbb{R}$$ ?

• Use $|\sin(x)|\le1$ and $\frac1x$ is decreasing on $(0,\infty)$ and $(-\infty,0)$ Commented Aug 21, 2022 at 2:58
• but $\frac{1}{x}$ is unbounded near $x=0$ Commented Aug 21, 2022 at 3:00
• @TymaGaidash Your comment does not lead to a proof of boundedness. Commented Aug 21, 2022 at 3:07
• You can use L'Hopital's rule to show that limit as $x \rightarrow 0$ is finite.
– Doug
Commented Aug 21, 2022 at 3:08
• @TymaGaidash It won't help near zero. Commented Aug 21, 2022 at 3:51

If you know that $$|\sin x| \le |x|$$ for all $$x$$, then the proof is easy: dividing both sides by $$|x$$|, we obtain $$\left| \frac{\sin x}{x} \right| \le 1$$ (for $$x \ne 0$$). Since $$f(x) = 1$$ by definition, this inequality holds for all $$x$$, which completes the proof.

So the question reduces to: how do you prove that $$|\sin x| \le |x|$$ for all $$x$$?

The answer to this question depends on how you are defining the sine function, and on what has already been proven about it: If you define $$\sin x$$ via a power series, the proof looks very different from if you define $$\sin x$$ as the inverse of the function $$g(x) = \int_0^x \frac{1}{\sqrt{1-t^2}}\, dt$$, which in turn looks very different from if you define $$\sin \theta$$ to be the $$y$$-coordinate of a point on the unit circle intercepted by the terminal side of an angle $$\theta$$ in standard position. Similarly, if you have already proven the fact that $$\frac{d}{dx}(\sin x) = \cos x$$ then you can base your argument off the fact that $$\sin x = \int_0^x \cos t \, dt$$ and $$|\cos t| \le 1$$ for all $$t$$. What counts as an acceptable proof is very context-dependent, and without knowing your context, it is impossible to say what is the right way to proceed.

Notice that

$$\lim_{x\to0}\frac{\sin x}{x}=1$$

(the proof of this is very standard, so I'll leave it out), and that

$$-\frac{1}{x}\leq\frac{\sin x}{x}\leq\frac{1}{x},$$

which yields that

$$\lim_{x\to\pm\infty}\frac{\sin x}{x}=0$$

by the squeeze theorem. Now let $$M>0$$, and use the above to find some $$R>0$$ such that $$\lvert f(x)\rvert\leq M$$ for all $$x\in(-\infty,-R)\cup(R,\infty)$$. Now notice that, as we has that the first limit was $$1$$, which is what $$f(0)$$ is defined as, $$f$$ is continuous, and so (as $$\lvert f\rvert$$ is then also continuous) $$\lvert f\rvert$$ attains a maximum value $$C$$ on the compact interval $$[-R,R]$$. Setting $$\alpha=\max\{C,M\}$$ we then clearly have that

$$\lvert f(x)\rvert\leq \alpha$$

for all $$x\in\mathbb{R}$$, hence $$f$$ is bounded.