Discussion
If \([x]\) is the greatest integer less than or equal to \(x\), then \(\displaystyle\lim_{x \to 0^{-}} \frac{\sin[x]}{[x]}\) is equal to
Answer: B
The limit is sin 1.
For \(-1 \le x \lt 0\), \([x] = -1\). So as \(x \to 0^{-}\) the expression is constant:
\[ \frac{\sin(-1)}{-1} = \frac{-\sin 1}{-1} = \sin 1 \]
The trap is option A: the standard limit \(\frac{\sin t}{t} \to 1\) needs \(t \to 0\), but here \([x]\) stays at −1. −sin 1 forgets that the denominator is also negative.
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