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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
Let f:(1,2)→Rf:(1,2)\to\mathbb{R} satisfies the inequality cos⁡(2x−4)−332<f(x)<x2∣4x−8∣x−2\dfrac{\cos(2x-4)-33}{2}<f(x)<\dfrac{x^{2}|4x-8|}{x-2}, ∀x∈(1,2)\forall x\in(1,2). Then lim⁡x→2−f(x)\displaystyle\lim_{x\to 2^{-}}f(x) is equal
A1616
B−16-16correct
CCannot be determined from the given information
DDoes not exist
Solution
Step 1: Compute the LHS bound as x→2−x\to 2^{-}:
lim⁡x→2−cos⁡(2x−4)−332=cos⁡0−332=1−332=−322=−16\lim_{x\to 2^{-}}\dfrac{\cos(2x-4)-33}{2}=\dfrac{\cos 0-33}{2}=\dfrac{1-33}{2}=\dfrac{-32}{2}=-16
 Step 2: Compute the RHS bound as x→2−x\to 2^{-}. For x<2x<2: 4x−8<04x-8<0, so ∣4x−8∣=8−4x=4(2−x)|4x-8|=8-4x=4(2-x). Also x−2<0x-2<0, so:
x2⋅4(2−x)x−2=x2⋅4(2−x)−(2−x)=−4x2\dfrac{x^{2}\cdot 4(2-x)}{x-2}=\dfrac{x^{2}\cdot 4(2-x)}{-(2-x)}=-4x^{2}
 Step 3: Take limit:
lim⁡x→2−(−4x2)=−16\lim_{x\to 2^{-}}(-4x^{2})=-16
 Step 4: Both bounds equal −16-16. By sandwich theorem:
lim⁡x→2−f(x)=−16\lim_{x\to 2^{-}}f(x)=-16
 Correct answer: (2) −16-16
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