If lim_{x} {rightarrow 0} left{ frac{1}{x^{8} | vedclass

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(B) The given expression is $\lim_{x}$ ${\rightarrow 0} \frac{1}{x^{8}} \left( 1 - \cos \frac{x^{2}}{2} - \cos \frac{x^{2}}{4} + \cos \frac{x^{2}}{2}

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The given expression is $\lim_{x}$ ${ightarrow 0} \frac{1}{x^{8}} \left( 1 - \cos \frac{x^{2}}{2} - \cos \frac{x^{2}}{4} + \cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4} ight)$.Factoring the numerator,we get $\lim_{x ightarrow 0} \frac{(1 - \cos \frac{x^{2}}{2})(1 - \cos \frac{x^{2}}{4})}{x^{8}}$.Using the limit formula $\lim_{	heta ightarrow 0} \frac{1 - \cos 	heta}{	heta^{2}} = \frac{1}{2}$,we rewrite the expression as:$\lim_{x}$ ${ightarrow 0} \left( \frac{1 - \cos \frac{x^{2}}{2}}{(x^{2}/2)^{2} \cdot 4} ight) \cdot \left( \frac{1 - \cos \frac{x^{2}}{4}}{(x^{2}/4)^{2} \cdot 16} ight) = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{16} = \frac{1}{256}$.Since $\frac{1}{256} = \frac{1}{2^{8}} = 2^{-8}$,we have $2^{-8} = 2^{-k}$.Therefore,$k = 8$.

If $\lim_{x}$ ${\rightarrow 0} \left\{ \frac{1}{x^{8}} \left( 1 - \cos \frac{x^{2}}{2} - \cos \frac{x^{2}}{4} + \cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4} \right) \right\} = 2^{-k}$,then the value of $k$ is

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