lim _{x} {rightarrow 0} 2left(frac{1-cos x sq | vedclass

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Question

(B) ધારો કે $L = \lim _{x \rightarrow 0} 2\left(\frac{1-\prod_{k=1}^{10} (\cos kx)^{1/k}}{x^2}\right)$.$\cos kx \approx 1 - \frac{(kx)^2}{2}$ વિસ્તરણન

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(B) ધારો કે $L = \lim _{x ightarrow 0} 2\left(\frac{1-\prod_{k=1}^{10} (\cos kx)^{1/k}}{x^2}ight)$.$\cos kx \approx 1 - \frac{(kx)^2}{2}$ વિસ્તરણનો ઉપયોગ કરતા,$(\cos kx)^{1/k} \approx (1 - \frac{k^2x^2}{2})^{1/k} \approx 1 - \frac{1}{k} \cdot \frac{k^2x^2}{2} = 1 - \frac{kx^2}{2}$.આમ,ગુણાકાર $\prod_{k=1}^{10} (\cos kx)^{1/k} \approx \prod_{k=1}^{10} (1 - \frac{kx^2}{2}) \approx 1 - \sum_{k=1}^{10} \frac{kx^2}{2}$.લિમિટમાં આ કિંમત મૂકતા:$L = \lim _{x ightarrow 0} 2\left(\frac{1 - (1 - \sum_{k=1}^{10} \frac{kx^2}{2})}{x^2}ight)$$L = \lim _{x ightarrow 0} 2\left(\frac{\sum_{k=1}^{10} \frac{kx^2}{2}}{x^2}ight) = \sum_{k=1}^{10} k$.$L = 1 + 2 + 3 + \ldots + 10 = \frac{10 	imes 11}{2} = 55$.

$\lim _{x}$ ${\rightarrow 0} 2\left(\frac{1-\cos x \sqrt{\cos 2 x} \sqrt[3]{\cos 3 x} \ldots \sqrt[10]{\cos 10 x}}{x^2}\right)$ ની કિંમત ............ છે.

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