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Question Number 186758 by EnterUsername last updated on 09/Feb/23

$$\mathrm{If}\left[t\right]\mathrm{denotes}\mathrm{the}\mathrm{integral}\mathrm{part}\mathrm{of}t,\mathrm{then}\underset{x\to 1}{\lim }\left[x\sin \pi x\right]$$$$\left(\mathrm{A}\right)\mathrm{equals}1\left(\mathrm{B}\right)\mathrm{equals}-1$$$$\left(\mathrm{C}\right)\mathrm{equals}0\left(\mathrm{D}\right)\mathrm{does}\mathrm{not}\mathrm{exist}$$

Answered by Gazella thomsonii last updated on 10/Feb/23

$$\underset{x\to 1}{\lim }\int x\sin \left(\pi x\right)\mathrm{d}x=\underset{x\to 1}{\lim }{\left(\frac{1}{\pi }\right)}^2\cdot (\sin \left(\pi x\right)-\pi x\cos \left(\pi x\right)$$$$\frac{1}{{\pi }^2}\left(\mathrm{D}\right)\cdot \cdot \cdot \cdot {\mathrm{Dosen}}^{\prime }\mathrm{t}\mathrm{Exist}$$

Answered by mr W last updated on 10/Feb/23

$$\underset{x\to 1^{-}}{\lim }\left[x\sin \pi x\right]=0$$$$\underset{x\to 1^{+}}{\lim }\left[x\sin \pi x\right]=-1$$$$\Rightarrow \underset{x\to 1}{\lim }\left[x\sin \pi x\right]{doesn}^{\prime }texist!$$

Commented by EnterUsername last updated on 10/Feb/23

Thank you, Sir.

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