Check if $\;f_y(x) = \dfrac{y \arctan(xy)}{y+1}\;$ converges uniformly when $y \to + \infty$ and $x \in [1, + \infty)$.
My work:
Firstly, we must find $\;\lim\limits_{y \to \infty} \dfrac{y \arctan(xy)}{y+1}\;$ and that limit is equal to $\,\dfrac{\pi}{2}\,$.
Then we are looking for $$\lim_{y \to \infty} \sup_{x \geq 1} \left| \frac{y \arctan(xy)}{y+1} - \frac{\pi}{2} \right|$$ If that limit is equal to $0$ then $f_y$ converges uniformly.
Let $\;\varphi(x) = \dfrac{y \arctan(xy)}{y+1} - \dfrac{\pi}{2}\,$. Easily we can see that $\varphi' > 0$ so that means that $\sup_{x\geq 1} \varphi$ is achieved when $x\to \infty$, which is equal to $$ \sup_{x \geq 1} \left|\frac{y \arctan(xy)}{y+1} - \frac{\pi}{2}\right|= \left| \frac{\frac{\pi}{2} y}{y+1} - \frac{\pi}{2}\right|$$ From that we get that $$\lim_{y \to \infty} \sup_{x \geq 1} \left| \frac{y \arctan(xy)}{y+1} - \frac{\pi}{2} \right|= 0$$ so it converges uniformly. But, result in my book is that it does not converges uniformly. Where I am wrong?
For $x\ge 1$ we have $$0\le {\pi\over 2}- f_y(x)={\pi \over 2}-{y\arctan xy\over y+1}\le {\pi\over 2}-{y\arctan y\over y+1}$$ Hence the limit is uniform.
If we extend the interval to $x>0$ then the uniform convergence fails as $${\pi\over 2}-f_y(y^{-1})={\pi\over 2}-{\pi\over 4}{y\over y+1}>{\pi \over 4}$$