Proving $\cot x =\alpha x$ has a solution $\forall \alpha>0$ in $(0,\frac\pi2)$ 
Prove $\cot x =\alpha x$ has a solution $\forall \alpha>0$ in the section $(0,\frac\pi2)$.

Well let's define: $g(x)=\cot x -\alpha x$
I know that $\cot$ goes to infinity as x go to zero, and go to zero as x go to $\pi/2$. Also, as $-\alpha x$ gets bigger the graph of $g(x)$ go below the $X$ axis. 
It goes through the $X$ axis without almost any dependency on  $\alpha$.
So I'm left with showing that the limits $\displaystyle\lim_{x\to0^+}g(x)=+\infty , \ \lim_{x\to{\frac\pi2}}g(x)=-\infty$
For minus infinity it's obvious since $\cot \frac\pi2=0$ but how can I show for sure that for $x\to0^+$ $\cot x > \alpha x$ ?
I know that $\cot x$ isn't always continuous but in this section it is (if I'm not mistaken).
 A: Well, you know that $$\lim_{x\searrow 0}\cot x=+\infty,$$ meaning that for any real $M,$ there is some $\delta>0$ such that $\cot x>M$ for $0<x<\delta.$ In particular, what can we conclude when we consider $M=\alpha\cdot\frac\pi2$?
As for your assertion that $$\lim_{x\nearrow\frac\pi2}g(x)=-\infty,$$ this is not the case. Rather, $$\lim_{x\nearrow\frac\pi2}g(x)=-\alpha\cdot\frac\pi2.$$ This, together with the above observation and Intermediate Value Theorem (since your function is continuous on the given interval, as you claim), will yield the desired result.
A: Hint: You only need to show that $\lim_{x\rightarrow \frac{\pi}2} g(x) < 0.$
A: This may not be what you want, but it is simple to understand.
Noting that $\cot x=\alpha x\iff\alpha=\frac{\cot x}{x}$, let us consider the latter. So, let $h(x)=\frac{\cot x}{x}=\frac{1}{x\tan x}.$
Since we get
$$h^\prime(x)=-\frac{\tan x+\frac{x}{\cos^2 x}}{x^2\tan^2 x}\lt0,$$
We know that this is a strict monotone decreasing function, and we know
$$\lim_{x\to 0+}h(x)=+\infty, \lim_{x\to \frac{\pi}{2}-}h(x)=0.$$
This "separation of a constant" method is sometimes helpful.
