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?

well, notice that $$\frac{x^2-2}{x^2-\sqrt{2}x-4} = \frac{(x-\sqrt{2})(x+\sqrt{2})}{(x-\sqrt{2})(x+2\sqrt{2})}$$ which gives the same answer as yours. 
