Discussion
\(\lim\limits_{x \to 0} \dfrac{x^{2}}{\sqrt{2} - \sqrt{1 + cosx}}\) is equal to
Answer: A
The limit is \(4\sqrt{2}\).
Rationalise the denominator:
\[ \begin{aligned} \sqrt{2} - \sqrt{1 + \cos x} &= \frac{1 - \cos x}{\sqrt{2} + \sqrt{1 + \cos x}} \\ \frac{x^{2}}{\sqrt{2} - \sqrt{1 + \cos x}} &= \frac{x^{2}}{1 - \cos x}\left(\sqrt{2} + \sqrt{1 + \cos x}\right) \end{aligned} \]
As \(x \to 0\), \(\frac{x^{2}}{1 - \cos x} \to 2\) and the bracket tends to \(2\sqrt{2}\), so the limit is \(2 \times 2\sqrt{2} = 4\sqrt{2}\). \(2\sqrt{2}\) forgets the factor 2 from \(1 - \cos x \approx \frac{x^{2}}{2}\).
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