Discussion
The value of \(\lim_{x \to 2^{+}} \frac{[x] - 2}{x - 2}\) is
Answer: E
The limit is 0.
As \(x \to 2^{+}\), x is slightly greater than 2, so \([x] = 2\) and the numerator is \(2 - 2 = 0\) for every such x.
\[ \begin{aligned} \lim_{x \to 2^{+}} \frac{[x] - 2}{x - 2} &= \lim_{x \to 2^{+}} \frac{0}{x - 2} \\ &= 0 \end{aligned} \]
The denominator is small but not zero, so the quotient is exactly 0 throughout. From the left the numerator would be \(1 - 2 = -1\) and the quotient would tend to infinity, which is why only the right-hand limit is asked.
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