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Question Number 74870 by vishalbhardwaj last updated on 02/Dec/19

$$\mathrm{solve}\mathrm{with}\mathrm{explanation}$$$$\mathrm{li}\underset{\mathrm{x}\to 0^{-}}{\mathrm{m}}\left[\frac{\mathrm{x}}{\mathrm{sinx}}\right],\mathrm{where}\left[\right]\mathrm{represents}\mathrm{greatest}\mathrm{integer}$$

Commented by mathmax by abdo last updated on 02/Dec/19

$$wehavesinx=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!}withradiusR=+\mathrm{\infty }$$$$sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-....\Rightarrow x-\frac{x^3}{6}⩽sinx⩽x(wecantake0⩽x⩽\frac{\pi }{2})$$$$1-\frac{x^2}{6}⩽\frac{sinx}{x}⩽1\Rightarrow [1-\frac{x^2}{6}]⩽\left[\frac{sinx}{x}\right]⩽1\Rightarrow 1+[-\frac{x^2}{2}]⩽\left[\frac{sinx}{x}\right]⩽1$$$$\Rightarrow {lim}_{x\to 0}\left[\frac{sinx}{x}\right]=1$$

Answered by mind is power last updated on 02/Dec/19

$$\frac{\mathrm{x}}{\sin \left(\mathrm{x}\right)}>0\mathrm{whe}\mathrm{x}\in ]-\frac{\pi }{2},0[$$$$\left[\frac{\mathrm{x}}{\sin \left(\mathrm{x}\right)}\right]⩽\frac{\mathrm{x}}{\sin \left(\mathrm{x}\right)}...\sqrt{\mathrm{E}}$$$$\mathrm{x}<\sin \left(\mathrm{x}\right),\mathrm{for}\mathrm{all}\mathrm{x}\in ]-\frac{\pi }{2},0[$$$$\mathrm{proof}$$$$\cos \left(\mathrm{t}\right)⩽1$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-\sin \left(\mathrm{x}\right)\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=1-\cos \left(\mathrm{x}\right)⩾0$$$$\mathrm{f}\mathrm{increase}\mathrm{f}\left(0\right)=0\Rightarrow \mathrm{x}-\sin \left(\mathrm{x}\right)<0\mathrm{\forall }\mathrm{x}\in [-\frac{\pi }{2},0[$$$$\Rightarrow \mathrm{x}<\sin \left(\mathrm{x}\right)$$$$\Rightarrow \frac{\mathrm{x}}{\sin \left(\mathrm{x}\right)}⩾1,\therefore \sin \left(\mathrm{x}\right)⩽0\therefore$$$$\Rightarrow \left[\frac{\mathrm{x}}{\sin \left(\mathrm{x}\right)}\right]⩾1$$$$\Rightarrow \mathrm{E}\iff 1⩽\left[\frac{\mathrm{x}}{\sin \left(\mathrm{x}\right)}\right]⩽\frac{\mathrm{x}}{\sin \left(\mathrm{x}\right)}$$$$\underset{x\to 0}{\lim }\frac{\mathrm{x}}{\sin \left(\mathrm{x}\right)}=\underset{x\to 0}{\lim }\frac{1}{\cos \left(\mathrm{x}\right)}=1\mathrm{hopitaks}\mathrm{Rulls}$$$$\Rightarrow \underset{x\to 0}{\lim }\left[\frac{\mathrm{x}}{\sin \left(\mathrm{x}\right)}\right]=1$$

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