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Question Number 176915 by thean last updated on 28/Sep/22

Answered by cortano1 last updated on 28/Sep/22

$$-1⩽\sin \mathrm{x}⩽1$$$$-1⩽\sin \mathrm{x}⩽1$$$$-\frac{\mathrm{x}+1}{{\mathrm{x}}^2+1}⩽\frac{(\mathrm{x}+1)\sin \mathrm{x}}{{\mathrm{x}}^2+1}⩽\frac{\mathrm{x}+1}{{\mathrm{x}}^2+1}$$$$\underset{x\to \mathrm{\infty }}{\lim }-\frac{\mathrm{x}+1}{{\mathrm{x}}^2+1}⩽\underset{x\to \mathrm{\infty }}{\lim }\frac{(\mathrm{x}+1)\sin \mathrm{x}}{{\mathrm{x}}^2+1}⩽\underset{x\to \mathrm{\infty }}{\lim }\frac{\mathrm{x}+1}{{\mathrm{x}}^2+1}$$$$\{\begin{array}{l}\underset{x\to \mathrm{\infty }}{\lim }-\frac{\mathrm{x}+1}{{\mathrm{x}}^2+1}=0\\ \underset{x\to \mathrm{\infty }}{\lim }\frac{\mathrm{x}+1}{{\mathrm{x}}^2+1}=0\end{array}$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }\frac{(\mathrm{x}+1)\sin \mathrm{x}}{{\mathrm{x}}^2+1}=0$$

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