Discussion
The value of \(\lim_{x \to 0} \frac{(x-sin2x)(2x-sinx)}{x^{2}}\) is equal to
Answer: D
The limit is -1.
Split the \(x^{2}\) between the two factors and use \(\frac{\sin kx}{x} \to k\):
\[ \begin{aligned} \frac{x - \sin 2x}{x} &\to 1 - 2 = -1 \\ \frac{2x - \sin x}{x} &\to 2 - 1 = 1 \end{aligned} \]
The product of the limits is \((-1)(1) = -1\). Taking \(\sin 2x\) as x instead of 2x near 0 would make the first factor 0 and give the trap answer 0. The values -2 and 2 come from multiplying the wrong pieces.
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