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Limits: Fractional Part Function

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

Question
lim⁡x→2+{x}sin⁡(x−2)(x−2)2\displaystyle\lim_{x\to 2^{+}}\dfrac{\{x\}\sin(x-2)}{(x-2)^{2}} where {⋅}\{\cdot\} is fractional part function
A00
B∞\infty
C11correct
DDoes not exist
Solution
Step 1: As x→2+x\to 2^{+}, {x}=x−2→0+\{x\}=x-2\to 0^{+}. Step 2: Substitute:
lim⁡x→2+(x−2)sin⁡(x−2)(x−2)2=lim⁡x→2+sin⁡(x−2)x−2=1\lim_{x\to 2^{+}}\dfrac{(x-2)\sin(x-2)}{(x-2)^{2}}=\lim_{x\to 2^{+}}\dfrac{\sin(x-2)}{x-2}=1
 Correct answer: (3) 11
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