jmc

We start by squaring both sides of the equation $$\begin{array}{rl}(\sqrt{2x}{)}^2& =(4x{)}^2\\ \Rightarrow 2x& =16x^2\\ \Rightarrow 16x^2-2x& =0\\ \Rightarrow 8x^2-x& =0\\ \Rightarrow x(8x-1)& =0\end{array}$$From here, we see that the two possible values of $x$ are $0$ and $\frac{1}{8}$. Since the problem only asks for the largest value of $x$, the final answer is $\boxed{\frac{1}{8}}$.