Help calculating a math limit Can anyone help with this limit?
\begin{equation*}
\lim_{x \rightarrow 4}
\frac{16\sqrt{x-\sqrt{x}}-3\sqrt{2}x-4\sqrt{2}}{16(x-4)^2}
\end{equation*}
I've tried a variable change of  \begin{equation*} y=\sqrt{x} \end{equation*} but this didn't help.

HINT: we have $$\frac{16\sqrt{x-\sqrt{x}}-\sqrt{2}(3x+4)}{16(x-4)^2}$$
multiply numerator and denominator by
$$16\sqrt{x-\sqrt{x}}+\sqrt{2}(3x+4)$$
the numerator is given after this multiplication as
$$-2 \left(\sqrt{x}-2\right)^2 \left(9 x+36 \sqrt{x}+4\right)$$
and write $$x-4$$ as $$(\sqrt{x}-2)(\sqrt{x}+2)$$
after all you will get this term
$$\frac{-2(\sqrt{x}-2)^2(9x+36\sqrt{x}+4)}{(\sqrt{x}-2)^2(\sqrt{x}+2)^2(16\sqrt{x-\sqrt{x}}+\sqrt{2}(3x+4))}$$
after cancelling $$(\sqrt{x}-2)^2$$ you can calculate the limit
