lim-n-4-n-1-cos-2-n-bemath-

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

\lim_{n \to \infty} 4^{n} (1 - \cos (\frac{α}{2^{n}})) ?

\sqrt{b e m a t h}

Answered by bobhans last updated on 06/Sep/20

\lim_{n \to \infty} 4^{n} (1 - \cos (\frac{α}{2^{n}})) = L

setting \frac{α}{2^{n}} = r; 2^{n} = \frac{α}{r} with r \to 0

L = \lim_{r \to 0} α^{2} (\frac{1 - \cos r}{r^{2}}) = α^{2} \lim_{r \to 0} \frac{2 \sin^{2} (\frac{r}{2})}{r^{2}}

L = \frac{α^{2}}{2} .

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

4^{n} (2 {s i n}^{2} \frac{α}{2^{n + 1}}) = 2^{2 n + 1} {(\frac{α}{2^{n + 1}})}^{2} = \frac{α^{2}}{2} s i n \frac{α}{2^{n}} \to \frac{α}{2^{n}}