Discussion
\(\lim_{x\to 0}\dfrac{\sqrt{cos^{2}x+3}-\sqrt{cos^{2}x+sinx+3}}{x}\) =
Answer: B
The limit is \(\frac{-1}{4}\).
Rationalise the numerator. The difference of the radicands is \(-\sin x\):
\[ \begin{aligned} &\frac{\sqrt{u} - \sqrt{u + \sin x}}{x} \\ &= \frac{-\sin x}{x\left(\sqrt{u} + \sqrt{u + \sin x}\right)} \end{aligned} \]
with \(u = \cos^{2}x + 3 \to 4\). As \(x \to 0\), \(\frac{\sin x}{x} \to 1\) and the bracket tends to \(2 + 2 = 4\), so the limit is \(-\frac{1}{4}\). \(\frac{1}{4}\) loses the sign, and \(\frac{-1}{2}\) forgets that two square roots of 4 are added in the denominator.
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