Discussion
The value of \(\lim_{x \to 0} \frac{\sqrt{1 + 3x} - \sqrt{1 - 3x}}{x}\) is
Answer: E
The limit is 3.
Rationalise the numerator.
\[ \begin{aligned} \frac{\sqrt{1 + 3x} - \sqrt{1 - 3x}}{x} &= \frac{(1 + 3x) - (1 - 3x)}{x(\sqrt{1 + 3x} + \sqrt{1 - 3x})} \\ &= \frac{6}{\sqrt{1 + 3x} + \sqrt{1 - 3x}} \to \frac{6}{2} = 3 \end{aligned} \]
Substituting x = 0 directly gives the indeterminate \(\frac{0}{0}\), not 0. \(\frac{1}{2}\) and \(\frac{1}{3}\) come from inverting part of the result.
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