Question

hallar el limite de la sgte.funcion
gracias :)​

Answer 1

$\displaystyle{\lim_{n\to\infty}\frac{1}{\cos(\frac{a}{2})\cos(\frac{a}{4})\cos(\frac{a}{8})...\cos(\frac{a}{2^n})}=L}$

si $a=0$ entonces el denominador es $1$ entonces $L=1$

si $a\neq 0$

$sen(a)=2cos(\frac{a}{2})sen(\frac{a}{2})=2^2cos(\frac{a}{2})cos(\frac{a}{4})sen(\frac{a}{4}=\\ 2^3cos(\frac{a}{2})cos(\frac{a}{4})cos(\frac{a}{8})sen(\frac{a}{8})=(...)=$

$2^n cos(\frac{a}{2})cos(\frac{a}{4})cos(\frac{a}{8})...cos(\frac{a}{2^n})sen(\frac{a}{2^n})\Rightarrow$

$cos(\frac{a}{2}).cos(\frac{a}{4})cos(\frac{a}{8})...cos(\frac{a}{2^n})=\frac{sen(a)}{2^nsen(\frac{a}{2^n})}$

$\mathrm{\large{Entonces:}}$

$\displaystyle{L=\lim_{n\to\infty}\frac{1}{\frac{sen(a)}{2^nsen(\frac{a}{2^n})}}=\lim_{x\to\infty}\frac{1}{sen(a)}\times 2^nsen(\frac{a}{2^n})}$

como la variable es $n$ entonces $\frac{1}{sen(a)}$ es constante, por propiedad de los límites sale multiplicando.

$\displaystyle{\frac{1}{sen(a)}\lim_{n\to\infty}2^n\sin(\frac{a}{2^n})=\frac{1}{sen(a)}\lim_{n\to\infty}\frac{sen(\frac{a}{2^n})}{\frac{1}{2^n}}}$

multiplicó y divido por $a$:

$\displaystyle{\frac{a}{sen(a)}\lim_{n\to\infty}\underbrace{\frac{sen(\frac{a}{2^n})}{\frac{a}{2^n}}_{1}}$

$\mathrm{\large{Entonces:}}$

$L=\frac{a}{sen(a)}$