Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
$\lim_{x \to c} f(x)$ does not exist when:
f(x) = |[x]| - |2x - 1|, c = 3
f(x) = [x] - x, c = 1
f(x) = [x]^2 - {-x}^2, c = 0
f(x) = tan(sgn x) / sgn x, c = 0
Step-by-Step Solution
Key Concept: Continuity at a point requires that $f(a^+) = f(a^-) = f(a)$ at the point of interest.
For option (A): $f(3^+) = 3 - 5 = -2$ and $f(3^-) = 2 - 4 = -2$, so the function is continuous at $x=3$. For option (B): $f(1^+) = 1 - 1 = 0$ and $f(1^-) = -1$, showing a jump discontinuity. For option (C): $f(0^+) = -1$ and $f(0^-) = 1$, indicating a discontinuity. For option (D): $f(0^+) = \tan 1$ and $f(0^-) = \tan 1$, confirming continuity. Options (A) and (C) are correct.
Correct Answer: 2,3