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JEE Main Mathematics question on Limits Continuity and Differentiability

Let f(x)=1−x(1+∣1−x∣)∣1−x∣cos⁡(11−x)f(x) = \dfrac{1 - x(1 + |1 - x|)}{|1 - x|} \cos\left(\dfrac{1}{1 - x}\right)

for x≠1x \neq 1 Then

(a) lim⁡x→1−f(x)=0\lim_{x \to 1^-} f(x) = 0

(b) lim⁡x→1−f(x)\lim_{x \to 1^-} f(x) does not exist

(c) lim⁡x→1+f(x)=0\lim_{x \to 1^+} f(x) = 0

(d) lim⁡x→1+f(x)\lim_{x \to 1^+} f(x) does not exist
  1. A.
    lim⁡x→1−f(x)=0\lim_{x \to 1^-} f(x) = 0
  2. B.
    lim⁡x→1−f(x)\lim_{x \to 1^-} f(x) does not exist
  3. C.
    lim⁡x→1+f(x)=0\lim_{x \to 1^+} f(x) = 0
  4. D.
    lim⁡x→1+f(x)\lim_{x \to 1^+} f(x) does not exist
Show answer and solution

Correct option: D

Solution

For x → 1⁺, let x = 1 + h where h → 0⁺. Then |1 - x| = h, and 1/(1-x) = -1/h → -∞.

f(x) = [1 - (1+h)(1+h)] / h · cos(-1/h) = [1 - (1+h)²] / h · cos(1/h)

[1 - (1+h)²] / h = (-2h - h²)/h = -(2 + h) → -2 as h → 0⁺.

So f(x) → -2 · cos(1/h) as h → 0⁺. Since cos(1/h) oscillates between -1 and 1 infinitely as h → 0⁺, the limit does not exist.

For x → 1⁻, let x = 1 - h where h → 0⁺. Then |1 - x| = h, and 1/(1-x) = 1/h → +∞.

f(x) = [1 - (1-h)(1+h)] / h · cos(1/h) = [1 - (1-h²)] / h · cos(1/h) = h²/h · cos(1/h) = h·cos(1/h) → 0.

So lim(x→1⁻) f(x) = 0 (options a and c are correct along with d, but the key answer is option 4: lim(x→1⁺) f(x) does not exist).

Previous-year question, JEE Main.

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