Verifying proof of $\lim_{x \to\sqrt{2}}\frac{x^2-2}{x^2+\sqrt{2}x-4} = \frac 2 3$ $$\lim_{x \to\sqrt{2}} \dfrac{x^2-2}{x^2+\sqrt{2}x-4} = \lim_{w \to2} \dfrac{w^2-4}{w^2+2w-8} =\lim_{w \to2} \dfrac{(w-2)(w+2)}{(w+4)(w-2)} = \frac 2 3$$
Change of variable:
$$w=\sqrt{2}x \Rightarrow x=\frac{w}{\sqrt{2}}\Rightarrow x^2=\frac{w^2}{2}\text{ and if }x \to \sqrt{2} \text{ then }w\to 2.$$
Is that correct?

That is correct, although you could do the same without the change of variables, and with some courage to manipulate the $\sqrt{2}$
$\lim\limits_{x\to \sqrt{2}} \frac{x^2 - 2 }{ x^2 + \sqrt{2}x - 4}= \lim\limits_{x\to \sqrt{2}} \frac{ ( x- \sqrt{2} ) ( x + \sqrt{2} ) }{ (x - \sqrt{2} ) ( x + 2 \sqrt{2} ) } = \lim\limits_{x\to \sqrt{2}} \frac{x + \sqrt{2}}{ x + 2\sqrt{2}} = \frac{2 \sqrt{2}}{3 \sqrt{2}} = \frac{2}{3}$
As I said, is just a question of courage, but irt may save you some time factoring the polynomials.
