Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade None
Question:
$\lim f(x)$ does not exist when (where $[x]$ denotes the greatest integer less than or equal to $x$)
$f(x) = [[x]] - [2x-1], c = 3$
$f(x) = [x] - x, c = 1$
$f(x) = {x}^2 - [{x}]^2, c = 0$
$f(x) = \frac{\tan(\sgn x)}{(\sgn x)}, c = 0$
Step-by-Step Solution
Key Concept: The limit exists only when RHL and LHL are equal; check discontinuity points carefully using the greatest integer function definition.
For option (A), evaluate $\lim_{x \to 3}([x] - [2x-1])$ by computing RHL = $\lim_{h \to 0}([3+h] - [6+2h-1]) = 3 - 5 = -2$ and LHL = $\lim_{h \to 0}([3-h] - [6-2h-1]) = 2 - 4 = -2$, so the limit exists and equals $-2$. For option (B), $\lim_{x \to 1}[x] - x$ has RHL = $0$ and LHL = $-1$, so the limit does not exist. For option (C), $\lim_{x \to 0}(x^2 - [-x]^2)$ has RHL = $\lim_{h \to 0}(h^2 - (1-h)^2) = -1$ and LHL = $\lim_{h \to 0}((1-h)^2 - h^2) = -1$, so the limit does not exist. For option (D), $\lim_{x \to 0}\frac{\tan(\sin x)}{\sin x}$ equals $\tan(1)$ from both sides.
Correct Answer: 2,3