Compute minimum of $1-\frac{\sin x}x$ over $|x|\ge1$

Question

Let $$h(x):=\left.\begin{cases}\displaystyle1-\frac{\sin x}x\;\;\;&\text{, if }x\ne0\\0&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x\in\mathbb R$$ and $$\alpha:=\inf_{\substack{x\in\mathbb R\\|x|\ge1}}h(x).$$

How can we compute $\alpha$?

If we take a look at the plot of $h$, the infimum should be attained at $x=1$:

How can we justify this rigorously?

Answer 1

We show that $h$ is strictly increasing on $(0, \pi/2)$, then show that for $x \ge \pi/2$, $h$ is bounded below.

For the first part, we take the derivative: for $x\in (0, \pi/2)$

$$h'(x) = -\frac {x\cos x - \sin x}{x^2} = \frac {\tan x- x}{x^2 \sec x} \ge 0$$

as $\tan x \ge x$ on $(0, \pi/2)$.

Now the lower bound of $h$ for $x \ge \pi/2$ is obtained by observing:

$$h(x) = 1 - \frac {\sin x}x \ge 1-\frac{1}x \ge 1-\frac {1}{\pi/2}$$

Combining the two results, we see that the infimum is obtained at $x=1$.