_ x → 0 ^ 2 x 2 - 1 + x equals

$\lim_{x \to 0}\dfrac{\sin^{2}x}{\sqrt{2} - \sqrt{1 + \cos x}}$

On Rationalisation, we get,

$= \lim_{x \to 0}\dfrac{\sin^{2}x}{\sqrt{2} - \sqrt{1 + \cos x}} \times \dfrac{\sqrt{2} + \sqrt{1 + \cos x}}{\sqrt{2} + \sqrt{1 + \cos x}}$

$= \lim_{x \to 0}\dfrac{{\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)}^{2}}{1 - \cos x} \times \left(\sqrt{2} + \sqrt{1 + \cos x}\right)$

$= \lim_{x \to 0}\dfrac{4\sin^{2}\dfrac{x}{2}\cos^{2}\dfrac{x}{2}}{2\sin^{2}\dfrac{x}{2}} \times \left(\sqrt{2} + \sqrt{1 + \cos x}\right)$

$= \lim_{x \to 0}2\cos^{2}\dfrac{x}{2}\left(\sqrt{2} + \sqrt{1 + \cos x}\right) = 2 \times 2\sqrt{2}$

$= 4\sqrt{2}$.