What methods/theorems are commonly used when trying to prove that a function has exactly one root within a given interval $(a,b)$, or that it has no roots?
I have the function $f(x)=\dfrac1x-\dfrac{\cos x}{\sin x}$. It is a strictly increasing function that is not defined for $a$ nor $b$....
Would just like a hint.
On
We have $f(x) = 1/x - \cos(x)/\sin(x)$. Note that, with the restriction of $x \in (0,\pi), x \neq \pi/2$, $$ 1/x - \cos(x)/\sin(x) = 0 \iff\\ \frac{\sin(x) - x\cos(x)}{x \sin(x)} = 0 \iff\\ \sin(x) - x\cos(x) = 0 \iff\\ \tan(x) - x = 0 $$ So, it suffices to show that $g(x) = \tan(x) - x$ has no roots on this interval. In order to do so, it suffices to note that $g$ is continuous and strictly increasing on the intervals where it's defined, and that we have the following limits: $$ \lim_{x \to 0^+}g(x) = 0\\ \lim_{x \to \pi/2^-}g(x) = \infty\\ \lim_{x \to \pi/2^+}g(x) = -\infty\\ \lim_{x \to \pi^-}g(x) < 0 $$
Using the expression of $f$ you gave in a comment above to explain a method:
$$\frac{1}{x}-\frac{\cos x}{\sin x} = \frac{\sin x - x\cos x}{x\sin x} = \frac{x-\frac{x^3}{6}-(x-\frac{x^3}{2}) +o(x^3)}{x\sin x} = \frac{\frac{2x^3}{6}+o(x^3)}{x^2+o(x^2)}\sim \frac{x}{3} $$ when $x\to 0$. Your function can thus be extended by continuity on $a=0$. Show you can also do that at $b=\pi$ (not even necessary in your case), and then apply the mean value theorem or your machinery of choice.