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Question Number 174152 by cortano1 last updated on 26/Jul/22

$$\underset{x\to 0}{\lim }\frac{1-\sin (\frac{\pi }{2}\cos x+2x)}{5x^2}=?$$

Answered by CElcedricjunior last updated on 26/Jul/22

$$\underset{x\to 0}{\lim }\frac{1-sin(\frac{\mathit{\pi }}{2}cosx+2x)}{5{\mathit{x}}^2}=\frac{0}{0}=\mathit{F}\mathbf{I}$$$$\mathrm{en}\mathrm{appliquant}{\mathrm{l}}^{\prime }\mathbf{hospital}\mathbf{on}\mathbf{a}$$$$\underset{x\to 0}{\lim }\frac{-(-\frac{\mathit{\pi }}{2}\mathbf{sinx}+2)\mathbf{cos}(\frac{\mathit{\pi }}{2}\mathbf{cosx}+2\mathbf{x})}{10\mathbf{x}}$$$$=\underset{x\to 0}{\lim }\frac{\left(\frac{\mathit{\pi }}{2}\mathbf{cosx}\right)\mathbf{cos}(\frac{\mathit{\pi }}{2}\mathbf{cosx}+2\mathbf{x})-(\frac{\mathit{\pi }}{2}\mathbf{sinx}-2)(-\frac{\mathit{\pi }}{2}\mathbf{sinx}+2)\mathbf{sin}(\frac{\mathit{\pi }}{2}\mathbf{cosx}+2\mathbf{x})}{10}$$$$=\frac{0+{(\frac{\mathit{\pi }}{2}\times 0-2)}^2\mathit{s}\mathit{i}\mathit{n}(\frac{\mathit{\pi }}{2}\times 1+2\times 0)}{10}$$$$=\frac{4}{10}=\frac{2}{5}$$$${\mathit{d}}^{\prime }o\stackrel{`}{u}$$$$\underset{x\to 0}{\lim }\frac{1-\sin (\frac{\pi }{2}\mathbf{cosx}+2\mathbf{x})}{5{\mathbf{x}}^2}=\frac{2}{5}$$$$.....lec\acute{el}\stackrel{`}{ebre}cedricjunior........$$

Answered by a.lgnaoui last updated on 26/Jul/22

$$\sin (\frac{\pi }{2}\cos \mathrm{x}+2\mathrm{x})=\sin \left(\frac{\pi }{2}\cos \mathrm{x}\right)\cos 2\mathrm{x}+\cos \left(\frac{\pi }{2}\cos \mathrm{x}\right)\sin 2\mathrm{x}$$$$\mathrm{L}={\lim }_{\mathrm{x}\to 0}\frac{1-[\sin \left(\frac{\pi }{2}\cos \mathrm{x}\right)\cos 2\mathrm{x}+\cos \left(\frac{\pi }{2}\cos \mathrm{x}\right)\sin 2\mathrm{x}]}{{5\mathrm{x}}^2}$$$$={\lim }_{\mathrm{x}\to 0}[\frac{1}{{5\mathrm{x}}^2}-\left(\frac{\frac{\pi }{2}+2\cos \frac{\pi }{2}\sin \mathrm{x}}{{5\mathrm{x}}^2}\right)]$$$${\lim }_{\mathrm{x}\to 0}[\frac{1}{{5\mathrm{x}}^2}-\frac{\frac{\pi }{2}}{{5\mathrm{x}}^2}];l{\mathrm{im}}_{\mathrm{x}\to 0}\frac{\sin \mathrm{x}}{\mathrm{x}}=1$$$$\mathrm{L}={\lim }_{\mathrm{x}\to 0}\left(\frac{2-\pi }{{10\mathrm{x}}^2}\right)=-\mathrm{\infty }$$

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