MHTCET 11 Apr 2026 Shift 1 Paper

Concept:The greatest integer function $[x]$ is constant between consecutive integers, so the two one-sided limits at $x = 3$ must be checked separately.Explanation:For $x \to 3^{-}$, $x$ takes values just less than 3, such as 2.9, 2.99, 2.999.Hence, $[x] = 2$.Substituting this value,$$\frac{[x]-3}{x-3}=\frac{2-3}{x-3}=\frac{-1}{x-3}$$Here, $x-3$ is a very small negative number, so $\frac{-1}{x-3} \to +\infty$.Therefore,$$\lim_{x \to 3^{-}}\frac{[x]-3}{x-3}=+\infty$$For $x \to 3^{+}$, $x$ takes values just greater than 3, such as 3.1, 3.01, 3.001.Hence, $[x] = 3$.Substituting this value,$$\frac{[x]-3}{x-3}=\frac{3-3}{x-3}=0$$Therefore,$$\lim_{x \to 3^{+}}\frac{[x]-3}{x-3}=0$$Since the left-hand limit and the right-hand limit are not equal, the two-sided limit does not exist.Answer:The limit does not exist.Correct option: D.