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Question 127 of 144

Q.Do the limits of following functions exist as x→0x \to 0? State reasons for your answer. sin⁡(x−⌊x⌋)x−⌊x⌋\dfrac{\sin(x - \lfloor x \rfloor)}{x - \lfloor x \rfloor}

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2020Subjective· 3mImportance★★★★★
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The floor function ⌊x⌋\lfloor x\rfloor jumps at x=0x=0, making the left and right limits of this expression genuinely different, so the overall limit fails to exist.

Consider g(x)=sin⁡(x−⌊x⌋)x−⌊x⌋g(x)=\dfrac{\sin(x-\lfloor x\rfloor)}{x-\lfloor x\rfloor} as x→0x\to0.

Right-hand limit (x→0+x\to0^+): for small positive xx, ⌊x⌋=0\lfloor x\rfloor=0, so x−⌊x⌋=x→0+x-\lfloor x\rfloor=x\to0^+. Then g(x)=sin⁡xx→1g(x)=\dfrac{\sin x}{x}\to1 (the standard small-angle limit).

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