Supremum and Infimum of x/sin x on [0, π/2]

The expression $x/\sin x$ is not defined at $x=0$, because both numerator and denominator are zero. Thus, taken literally, it is a function on $(0,\pi/2]$, not on the entire closed interval $[0,\pi/2]$. On its actual domain, its infimum is $1$ and its supremum is $\pi/2$. The infimum is not attained, while the supremum is.

If the intended function is the continuous extension at zero, set $f(0)=1$. Then the minimum is also $1$, and the maximum remains $\pi/2$.

Show that the quotient is increasing

For $0<x\leq \pi/2$, differentiate using the quotient rule:

$$f'(x)=\frac{\sin x-x\cos x}{\sin^2 x}.$$

The denominator is positive on this domain, so it is enough to show that the numerator is positive. Define

$$h(x)=\sin x-x\cos x.$$

Since $h(0)=0$ and

$$h'(x)=x\sin x>0 \qquad (0<x\leq \pi/2),$$

$h$ is strictly increasing from zero. Therefore $h(x)>0$ for every $x>0$, and hence $f'(x)>0$. This proves that $x/\sin x$ is strictly increasing on $(0,\pi/2]$.

Find the infimum and supremum

Because the function is increasing, its smallest values occur near the left endpoint and its largest value occurs at the right endpoint.

At the right endpoint,

$$f\left(\frac{\pi}{2}\right)=\frac{\pi/2}{\sin(\pi/2)}=\frac{\pi}{2}.$$

So $\pi/2$ is attained and is the maximum, and therefore the supremum.

At the left endpoint, take the limit:

$$\lim_{x\to 0^+}\frac{x}{\sin x}=1,$$

using the standard limit $\lim_{x\to 0}\sin x/x=1$. Since the function is strictly increasing, every value on its domain is greater than $1$. Values can get arbitrarily close to $1$ as $x$ approaches zero, but no allowed input gives the value $1$. Thus $1$ is the infimum, not a minimum.

Equivalently, the range on the literal domain is $(1,\pi/2]$. If you extend the function continuously by setting $f(0)=1$, its range becomes $[1,\pi/2]$.

A concrete check

At $x=\pi/6$, the value is

$$f\left(\frac{\pi}{6}\right)=\frac{\pi/6}{\sin(\pi/6)}=\frac{\pi/6}{1/2}=\frac{\pi}{3}.$$

Numerically, $1<\pi/3<\pi/2$, which is consistent with the increasing behavior and the endpoint bounds. The limit near zero is $1$, but it is not obtained by substituting zero into the original quotient.

Common mistakes

• Substituting $x=0$ directly: This gives $0/0$, which is undefined. A limit can describe the behavior near zero, but it does not assign a value at zero unless you explicitly define an extension.
• Calling the infimum a minimum: For the domain $(0,\pi/2]$, the value $1$ is approached but never reached. A minimum must be an actual function value.
• Confusing supremum with maximum: Here they agree because the right endpoint is in the domain and the function takes the value $\pi/2$ there.
• Using monotonicity without checking the domain: The derivative argument applies where $\sin x\neq 0$; in the interval under discussion, that means $x>0$.