Function $\frac{\sin x}{x}$

Question

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 $(-\infty,0)$ — Тyma Gaidash, Aug 21 '22 at 02:58
@TymaGaidash Your comment does not lead to a proof of boundedness. — coffeemath, Aug 21 '22 at 03:07
Think about how $\sin(x)/x$ behave when $x$ is near to zero, in limit terms. — MrSelberg, Aug 21 '22 at 03:08
You can use L'Hopital's rule to show that limit as $x \rightarrow 0$ is finite. — , Aug 21 '22 at 03:08
Let $f:x\mapsto \sin(x)/x$. Technically speaking, $f(0)$ is not defined. But, in practice, we can define the function $$\operatorname{sinc}(x)=\begin{cases}\sin(x)/x & x\neq 0 \ 1 & x=0\end{cases}$$ Which has no singularity at $x=0$. — K.defaoite, Aug 21 '22 at 03:30
@Doug Don't abuse L'Hôpital's rule when it's just the definition of the derivative. — Ted Shifrin, Aug 21 '22 at 03:31
@mweiss i hope you understand that there is a problem for $x=0$ — blessed, Aug 21 '22 at 03:31
@blessed Yes, of course, but from $|\sin x| < |x|$ you can easily prove that your function $f$ satisfies $|f(x)| \le 1$ for all $x$. — mweiss, Aug 21 '22 at 03:34
using $\lim_{x\to0}\frac{sin x}{x}=1$, the function is continuous at $x=0$, which gives boundedness in $(-\delta,\delta)$ for some $\delta>0$. For the remaining part, obviously the function is bounded. — blessed, Aug 21 '22 at 03:37
@coffeemath The goal of the comment was to give information to help with the proof. If $\frac1x$ is decreasing, then it will help show that $\frac{\sin(x)}x$ is bounded. — Тyma Gaidash, Aug 21 '22 at 03:48

score 2 · Answer 1 · answered Aug 21 '22 at 13:31

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.

score 0 · Answer 2 · answered Aug 21 '22 at 13:41

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.

Function $\frac{\sin x}{x}$

2 Answers2