If mathop {lim }limits_{n to infty } frac{1}{ | vedclass

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Question

(A) Given $\mathop {\lim }\limits_{n 	o \infty } \frac{1}{{10 + {{\left( {2\cos x} ight)}^{2n}}}} = 0$. For the limit to be $0$,the denominator mus

Vedclass · Accepted Answer

$[0, \frac{\sqrt{3}}{2})$

Vedclass · Answer

(A) Given $\mathop {\lim }\limits_{n 	o \infty } \frac{1}{{10 + {{\left( {2\cos x} ight)}^{2n}}}} = 0$. For the limit to be $0$,the denominator must approach $\infty$ as $n 	o \infty$. This happens if $|2\cos x| > 1$. $|\cos x| > \frac{1}{2}$. Since $\sin^2 x = 1 - \cos^2 x$,we have $\cos^2 x > \frac{1}{4}$. $1 - \sin^2 x > \frac{1}{4} \Rightarrow \sin^2 x $|\sin x| Also,$|\sin x| \ge 0$. Thus,the set of values is $[0, \frac{\sqrt{3}}{2})$.

If $\mathop {\lim }\limits_{n \to \infty } \frac{1}{{10 + {{\left( {2\cos x} \right)}^{2n}}}} = 0$,then the complete set of all possible values of $|\sin x|$ is:

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