Question

The value of $\lim_{x\rightarrow 2^{+}}\frac{[x]-2}{x-2}$ is

• A. -2
• B. 4
• C. 2
• D. 1
• E. 0

Hint:

Logic Tip: Do not confuse exactly zero divided by a tiny number with the indeterminate form $0/0$. Here, the numerator is absolute $0$ *before* the limit operation is fully executed, making the entire fraction exactly $0$ at every point arbitrarily close to 2.

Answer 1

Answer: (E)

Concept:
The function $[x]$ denotes the greatest integer less than or equal to $x$ (the floor function). When calculating right-hand limits ($x \rightarrow a^+$), we consider values of $x$ that are infinitesimally larger than $a$.
Step 1: Evaluate the greatest integer function near the limit.
We are finding the limit as $x \rightarrow 2^+$. This means $x$ is approaching 2 from the right side (e.g., $x = 2.0001$). For any value of $x$ strictly between 2 and 3 ($2 \le x<3$), the greatest integer function yields exactly 2: $$[x] = 2$$
Step 2: Substitute this into the limit expression.
Replace $[x]$ with its exact value in the neighborhood of $2^+$: $$\lim_{x\rightarrow 2^{+}} \frac{[x]-2}{x-2} = \lim_{x\rightarrow 2^{+}} \frac{2-2}{x-2}$$
Step 3: Evaluate the simplified limit.
The numerator evaluates exactly to $0$. The denominator $x-2$ is approaching $0$ but is strictly positive and non-zero (since $x>2$). $$= \lim_{x\rightarrow 2^{+}} \frac{0}{x-2} = 0$$