Q58 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
The value of $\lim_{x \to 3} \dfrac{[x] - 3}{x - 3}$, where $[\cdot]$ denotes the greatest integer function, is.....
- A$\infty$
- B$1$
- C$0$
- Correct answerDdoes not exist
Solution
Evaluate the left-hand limit (LHL) and right-hand limit (RHL) at $x = 3$.
For LHL, let $x \to 3^{-}$. Then $[x] = 2$.
LHL $= \lim_{x \to 3^{-}} \dfrac{[x] - 3}{x - 3} = \lim_{x \to 3^{-}} \dfrac{2 - 3}{x - 3} = \lim_{x \to 3^{-}} \dfrac{-1}{x - 3} = \infty$
For RHL, let $x \to 3^{+}$. Then $[x] = 3$.
RHL $= \lim_{x \to 3^{+}} \dfrac{[x] - 3}{x - 3} = \lim_{x \to 3^{+}} \dfrac{3 - 3}{x - 3} = \lim_{x \to 3^{+}} 0 = 0$
Since LHL $\neq$ RHL, the limit does not exist.
Answer: does not exist
For LHL, let $x \to 3^{-}$. Then $[x] = 2$.
LHL $= \lim_{x \to 3^{-}} \dfrac{[x] - 3}{x - 3} = \lim_{x \to 3^{-}} \dfrac{2 - 3}{x - 3} = \lim_{x \to 3^{-}} \dfrac{-1}{x - 3} = \infty$
For RHL, let $x \to 3^{+}$. Then $[x] = 3$.
RHL $= \lim_{x \to 3^{+}} \dfrac{[x] - 3}{x - 3} = \lim_{x \to 3^{+}} \dfrac{3 - 3}{x - 3} = \lim_{x \to 3^{+}} 0 = 0$
Since LHL $\neq$ RHL, the limit does not exist.
Answer: does not exist
Hence, the correct answer is option: 'D'.