Discussion
Let \([a]\) be the greatest integer less than or equal to a, then \(\lim_{x \to 0^{+}} x\Big\{\Big[\frac{1}{x}\Big]+\Big[\frac{2}{x}\Big]\Big\}\) is equal to
Answer: C
The limit is 3.
Use \(t - 1 \lt [t] \le t\). With \(t = \frac{1}{x}\) and \(t = \frac{2}{x}\) for \(x \gt 0\):
\[ \begin{aligned} x\left(\frac{3}{x} - 2\right) &\lt x\left\{\left[\frac{1}{x}\right] + \left[\frac{2}{x}\right]\right\} \\ &\le x \cdot \frac{3}{x} \end{aligned} \]
The left side is \(3 - 2x \to 3\) and the right side is 3, so by the squeeze theorem the limit is 3. Taking only one of the two terms gives 1 or 2, and 0 treats the brackets as bounded.
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