Category: Limits

Question Number 174876 by infinityaction last updated on 13/Aug/22 $$\mathrm{let}f:[0,1]\to \mathbb{R}bea\mathrm{continuous}$$$$\mathrm{function}\mathrm{ditermine}(\mathrm{with}\mathrm{appropriate}$$$$\mathrm{justification})\mathrm{the}\mathrm{following}$$$$\mathrm{limit}:\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^1{nx}^{n}f\left(x\right)dx$$ Answered by TheHoneyCat…

Question Number 174841 by infinityaction last updated on 12/Aug/22 $$\underset{x\to 0}{\lim }\frac{x\sin \left(\sin x\right)-{\sin }^{2}x}{x^6}$$ Answered by Ar Brandon last updated on 12/Aug/22 $$\mathcal{A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\mathrm{sin}\left(\mathrm{sin}{x}\right)−\mathrm{sin}^{\mathrm{2}}…

Question Number 109284 by bobhans last updated on 22/Aug/20 Answered by abdomsup last updated on 22/Aug/20 $$f\left(x\right)=\frac{\sqrt{\pi -2x}-\sqrt{\pi +4x}}{cos(x-\frac{\pi }{2})}$$$$wehavef\left(x\right)=\frac{\sqrt{\pi -2x}-\sqrt{\pi +4x}}{sinx}$$$$\sqrt{\pi -2x}=\sqrt{\pi }\sqrt{1-\frac{2x}{\pi }}\sim \sqrt{\pi }(1-\frac{x}{\pi })$$$$\sqrt{\pi +4x}=\sqrt{\pi }\sqrt{1+\frac{4x}{\pi }}\sim \sqrt{\pi }(1+\frac{2x}{\pi })$$$${sinx}\:\sim{x}\:\Rightarrow…

Question Number 174715 by infinityaction last updated on 09/Aug/22 $$\underset{x\to \mathrm{\infty }}{\lim cos}\left(\frac{\pi }{4}\right)\cos \left(\frac{\pi }{8}\right)\dots \cos \left(\frac{\pi }{2^{n+1}}\right)$$ Answered by abdullahoudou last updated on 09/Aug/22 $$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}^{{n}−\mathrm{1}} }\right)\frac{\mathrm{1}}{\mathrm{2sin}\:\left(\pi/\mathrm{2}^{{n}} \right)}…

Question Number 109139 by bemath last updated on 21/Aug/20 $$\underset{x\to -\mathrm{\infty }}{\lim }(1-x)\sin \mid 2x-2\mid ?$$ Commented by kaivan.ahmadi last updated on 21/Aug/20 $$x\to -\mathrm{\infty }\Rightarrow 2x-2\to -\mathrm{\infty }\Rightarrow \mid 2x-2\mid \to +\mathrm{\infty }\Rightarrow$$$$-1⩽sin\mid 2x-2\mid ⩽1isanelementofRwe$$$${say}\:{a}.…