Discussion
The value of \(\displaystyle \lim_{x \to 0} \frac{x - \tan(3x)}{\sin(2x)}\) is equal to
Answer: E
The limit is \(-1\).
Divide the numerator and denominator by \(x\) and use \(\frac{\tan kx}{x} \to k\) and \(\frac{\sin kx}{x} \to k\) as \(x \to 0\):
\[ \begin{aligned} \lim_{x \to 0} \frac{1 - \frac{\tan 3x}{x}}{\frac{\sin 2x}{x}} &= \frac{1 - 3}{2} \\ &= -1 \end{aligned} \]
The option \(-\frac{3}{2}\) ignores the \(x\) in the numerator, and \(-2\) forgets to divide by the 2 from \(\sin 2x\).
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