UPSC NDA I 2025 Mathematics Paper

Concept:The limit exists only when the left-hand and right-hand limits are equal. Here, we analyze the behavior of $\sin(\lfloor x \rfloor)$ and $|x|$ near zero.Explanation:For $x \to 0^+$, $\lfloor x \rfloor = 0$, so $f(x) = \sin 0 = 0$. Also $g(x) = x$. Then $\frac{f(x)}{g(x)} = \frac{0}{x} = 0$. So right-hand limit is $0$.For $x \to 0^-$, $\lfloor x \rfloor = -1$, so $f(x) = \sin(-1) = -\sin 1$, a non‑zero constant. Also $g(x) = -x$. Then $\frac{f(x)}{g(x)} = \frac{-\sin 1}{-x} = \frac{\sin 1}{x}$. As $x \to 0^-$, this ratio tends to $-\infty$ (or is undefined because denominator goes to $0$ while numerator is non‑zero).Since the left‑hand limit (does not exist/finite) and right‑hand limit ($0$) are not equal, the overall limit does not exist.Answer:D. Limit does not exist