How to prove $f(x,y) = \frac{2 \log(y) \sqrt{y ( x-y)}}{x \log(x)}$ is bounded by $1$ for $(x,y) \to (0,0) $? Let
$$f(x,y) = \frac{2 \log(y) \sqrt{y ( x-y)}}{x \log(x)} $$
with $(x,y) \in D = \{(x,y)\mid 0 < y \leq x \le 1 \}$.
How would one show (or disprove)  that
$$
\forall \epsilon > 0\ \ \exists \delta>0 \ \ \forall (x,y) \in D \cap B_\delta(0) \quad f(x,y) < 1 + \epsilon
$$
where $B_\delta(0)$ is the disk with radius $\delta$ and center $(0,0)$?
Please note that this is not just a bound on the limit of the function, as it is not well-defined. Choosing another coordinate $0 \le \theta \le 1$ with $y = \theta x$ we get
$$
     f(x, \theta x) = \frac{2 (\log(\theta) + \log(x)) \sqrt{\theta (1-\theta)}}{\log(x)}\\
$$
such that
$$
\lim_{x \to 0} f(x, \theta x) = 2 \sqrt{\theta (1-\theta)}.
$$
So the limit is not well-defined as it depends on how the origin is approached.
Also, although the above limit $2 \sqrt{\theta (1-\theta)}$ is indeed bound by 1, it does not directly prove the original question, because $\theta$ is assumed constant with respect to $x$. More specifically, this limit shows (variable ranges omitted for clarity)
$$
\forall \epsilon \ \ \forall \theta\ \ \exists \delta \ \ \forall x < \delta \quad f(x,\theta x) < 1 + \epsilon
$$
while the original problem corresponds to the stronger (note the different quantifier ordering):
$$
\forall \epsilon \ \ \exists \delta \ \ \forall \theta\ \ \forall x < \delta \quad f(x,\theta x) < 1 + \epsilon
$$
 A: In polar coordinates:
$$f(r,\theta) = 
\frac{2 \log(r\sin \theta) \sqrt{\sin \theta (\cos \theta-\sin \theta)}}{\cos \theta \log(r\cos \theta)}$$
You can show that the function:
$$2\sqrt{\sin \theta (\cos \theta-\sin \theta)} \sec (\theta) <1$$
Simply by taking the derivative and showing that it obtains the maximum value $1$ at $\theta\to\tan^{-1}(1/2)$. So:
$$f(r,\theta)< \frac{\log(r\sin \theta)}{\log(r\cos \theta)} =
\frac{\log(r)+\log(\sin \theta)}{\log(r)+\log(\cos \theta)}\sim
1+\frac{\sin \theta - \cos \theta}{\log(r)}+O(\log^{-2}(r))$$
Clearly, for small enough $r\to\delta>0$,
$$\epsilon = \frac{\sin \theta - \cos \theta}{\log(\delta)} > 0 $$.
A: $$ 
\begin{eqnarray}
f(x, \theta x) &=& \frac{2 (\log(\theta) + \log(x)) \sqrt{\theta (1-\theta)}}{\log(x)}\\
 &=& \frac{- 2 \log(\theta) \sqrt{\theta (1-\theta)}}{-\log(x)} + 2 \sqrt{\theta (1-\theta)} \\
&\le&\frac{- 2 \log(\theta) \sqrt{\theta (1-\theta)}}{- \log(x)} + 1 \\ 
&\le&\frac{V^*}{-\log(x)} + 1 \\ 
\end{eqnarray}
$$
with $V^* = \max_{\theta \in [0,1]} (- 2 \log(\theta) \sqrt{\theta (1-\theta)}) \approx 1.3818$ (at $\theta^* \approx 0.1041$).
So 
$$
\begin{eqnarray}
f(x, \theta x) < 1 + \epsilon & \Leftarrow  & \frac{V^*}{-\log(x)} + 1 < 1 + \epsilon \\&\Leftrightarrow & x < e^{-V^*/\epsilon}
\end{eqnarray}
$$
So with $\delta(\epsilon)=e^{-V^*/\epsilon}$ we have shown that
$$
\forall \epsilon > 0 \quad \forall x \in\ ]0, \delta(\epsilon)[\quad \forall \theta \in [0,1]\quad f(x, \theta x) < 1 + \epsilon.
$$
