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Question Number 112083 by bemath last updated on 06/Sep/20

$$\underset{n\to \mathrm{\infty }}{\lim }{4}^n(1-\cos \left(\frac{\alpha }{2^n}\right))?$$$$\sqrt{bemath}$$

Answered by bobhans last updated on 06/Sep/20

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{4}^{\mathrm{n}}(1-\cos \left(\frac{\alpha }{2^{\mathrm{n}}}\right))=\mathrm{L}$$$$\mathrm{setting}\frac{\alpha }{2^{\mathrm{n}}}=\mathrm{r};2^{\mathrm{n}}=\frac{\alpha }{\mathrm{r}}\mathrm{with}\mathrm{r}\to 0$$$$\mathrm{L}=\underset{\mathrm{r}\to 0}{\lim }{\alpha }^2\left(\frac{1-\cos \mathrm{r}}{{\mathrm{r}}^2}\right)={\alpha }^2\underset{\mathrm{r}\to 0}{\lim }\frac{2\sin {}^2\left(\frac{\mathrm{r}}{2}\right)}{{\mathrm{r}}^2}$$$$\mathrm{L}=\frac{{\alpha }^2}{2}.$$

Answered by Dwaipayan Shikari last updated on 06/Sep/20

$$4^n\left(2{sin}^2\frac{\alpha }{2^{n+1}}\right)=2^{2n+1}{\left(\frac{\alpha }{2^{n+1}}\right)}^2=\frac{{\alpha }^2}{2}sin\frac{\alpha }{2^n}\to \frac{\alpha }{2^n}$$

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