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Question Number 37817 by prof Abdo imad last updated on 17/Jun/18

$$calculate{lim}_{n\to +\mathrm{\infty }}{x}^n(1-cos\left(\frac{\pi }{x^n}\right))withx$$$$fromRandx\ne 0$$

Commented by prof Abdo imad last updated on 24/Jun/18

$$if\mid x\mid <1{lim}_{n\to +\mathrm{\infty }}{x}^n=0chang.x^n=\frac{1}{t}$$$${lim}_{n\to +\mathrm{\infty }}{x}^n(1-cos\left(\frac{\pi }{x^n}\right))={lim}_{t\to +\mathrm{\infty }}\frac{1}{t}(1-cos\left(\pi t\right))$$$$={lim}_{t\to +\mathrm{\infty }}\frac{2{sin}^2\left(\frac{\pi t}{2}\right)}{t}=0because0⩽{sin}^2\left(\frac{\pi t}{2}\right)⩽1$$$$ifx>1{lim}_{n\to +\mathrm{\infty }}{x}^n=+\mathrm{\infty }chang.x^n=\frac{1}{t}$$$${lim}_{n\to +\mathrm{\infty }}{x}^n(1-cos\left(\frac{\pi }{x^n}\right))={lim}_{t\to 0}\frac{1}{t}(1-cos\left(\pi t\right))$$$$but1-cos\left(\pi t\right)\sim \frac{{\pi }^2{t}^2}{2}(t\to 0)\Rightarrow$$$$\frac{1-cos\left(\pi t\right)}{t}\sim \frac{{\pi }^{2}t}{2}\to 0(t\to 0)so$$$${lim}_{n\to +\mathrm{\infty }}{x}^n(1-cos\left(\frac{\pi }{x^n}\right))=0.$$

Answered by behi83417@gmail.com last updated on 17/Jun/18

$$\frac{1}{x^n}=t,x\to +\mathrm{\infty }\Rightarrow t\to 0$$$$L=\underset{x\to 0}{\lim }\frac{1-cos\pi t}{t}=\underset{x\to 0}{\lim }\frac{\frac{{\left(\pi t\right)}^2}{2}}{t}=0.◼$$

Commented by math khazana by abdo last updated on 18/Jun/18

$$sirBehibecarrefuln\to +\mathrm{\infty }notx...$$

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