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?
 A: 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.
A: 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. 
A: Your work is correct and helps in removing square roots. But you can dispense with the change of variables.
Let $f(x)=x^2+x\sqrt{2}-4$; then $f(\sqrt{2})=0$, so you know $\sqrt{2}$ is a root of the polynomial. Then the usual scheme (you may be used to a different one)
$$
\begin{array}{r|rr|r}
& 1 & \sqrt{2} & -4 \\
\sqrt{2} & & \sqrt{2} & 4\\
\hline
& 1 & 2\sqrt{2} & 0
\end{array}
$$
tells you $x^2+x\sqrt{2}-4=(x-\sqrt{2})(x+2\sqrt{2})$. Similarly, $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$, so you can factor out $x-\sqrt{2}$ from the numerator and denominator.
Alternatively, use l'Hôpital's rule (maybe for checking if you're not officially allowed to use it):
$$
\lim_{x \to\sqrt{2}}
\dfrac{x^{2}-2}{x^{2}+\sqrt{2}x-4}
\overset{\text{(H)}}{=}
\lim_{x\to\sqrt{2}}
\dfrac{2x}{2x+\sqrt{2}}=\frac{2}{3}
$$
