Show that $\frac{1}{x}-\sin(x)$ has exactly one root in the interval $(0,\frac{\pi}{2}]$

Question

I have problems showing, that this function has exact one root in the interval $\left(0,\frac{\pi}{2}\right]$: $$f(x):=\frac{1}{x}-\sin(x)$$

My idea was to use the Intermediate value theorem, but the problem is that the theorem is only specified for closed intervals $[a,b]$ and that the function must be continuous in that interval - which both is not the case. The limits are $$\lim\limits_{x\rightarrow 0} = -\infty$$ and $$\lim\limits_{x\rightarrow \frac{\pi}{2}} = \frac{2}{\pi}-1$$

I plotted the function with WolframAlpha and it showed that the root must be at around 1.1...(the value is irrelvant in this case).

My next idea was to calculate the derivate to show if the function is monotonic increasing/decreasing, which solves to: $$f'(x)=-\frac{1}{x^2}-cos(x) <0$$ => monotonically decreasing. But I still can't say anything about the root.

But I am stuck, I don't know if my idea is right and/or how to continue, especially because the interval is half open. Any ideas on how to solve this?

Answer 1

Note that

$$\lim_{x\to 0^+} f(x)=+\infty.$$

As you say

$$\lim_{x\to (\pi/2)^-} f(x)=\frac{2}{\pi}-1<0.$$

Because of continuity of $f$ there exists $a,b\in (0,\pi/2)$ such that $a<b,$ $f(a)>0$ and $f(b)<0.$ Then, by Bolzano's theorem, there exists $c\in (a,b)$ such that $f(c)=0.$

Now assume there are two solutions of the equation, say $c,d$ with $c<d.$ Then, by Rolle's theorem there exists $e\in (c,d)$ such that $f'(e)=0.$ This contradicts the fact that $f'<0.$ Thus, the initial assumption is not correct. That is, $f$ has no two zeros.

Answer 2

Hints:

Note that, for $x \leq 1$, $\dfrac1x - \sin x > 0$ so you can restrict your attention to $[1,\pi/2]$.

Next, assume that there are two zeros. Now apply the mean value theorem (rather than intermediate value theorem).