Discussion
Let \([x]\) denote the greatest integer less than or equal to \(x\) . Then \(\displaystyle\lim_{x \to 0^{-}} \frac{\sin[x]}{[x]}\) is equal to
Answer: E
The limit is sin1.
As \(x \to 0^{-}\), \(x\) is a small negative number, so \([x] = -1\) throughout the approach. The expression is constant there:
\[ \begin{aligned} \frac{\sin[x]}{[x]} &= \frac{\sin(-1)}{-1} \\ &= \frac{-\sin 1}{-1} \\ &= \sin 1 \end{aligned} \]
The trap is to use \(\frac{\sin\theta}{\theta} \to 1\), which needs \(\theta \to 0\); here \([x]\) stays at -1 and never approaches 0. Forgetting that the two minus signs cancel gives -sin1.
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