1985

JEE Main 1985 Mathematics question on Limits Continuity and Differentiability

If f(x)={sin⁡[x][x],[x]≠00,[x]=0f(x) = \begin{cases} \frac{\sin[x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{cases}

where, [x][x] denotes the greatest integer less than or equal to xx, then lim⁡x→0f(x)\lim_{x \to 0} f(x) equals \hspace{2cm}
  1. A.
    (a) 1
  2. B.
    (b) 0
  3. C.
    (c) −1-1
  4. D.
    (d) None of these
Show answer and solution

Correct option: D

Solution

For x → 0⁺, [x] = 0, so f(x) = 0, giving right-hand limit = 0.

For x → 0⁻, [x] = -1, so f(x) = sin(-1)/(-1) = sin(1), giving left-hand limit = sin(1).

Since sin(1) ≠ 0, the left-hand and right-hand limits are unequal, so the limit does not exist — which is none of the values listed (1, 0, or -1).

[1985]

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