lim _{x} {rightarrow 0} frac{8}{x^8}left[1-co | vedclass

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(D) ધારો કે $L = \lim _{x}$ ${\rightarrow 0} \frac{8}{x^8}\left[1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2

Vedclass · Accepted Answer

$\frac{1}{32}$

Vedclass · Answer

(D) ધારો કે $L = \lim _{x}$ ${ightarrow 0} \frac{8}{x^8}\left[1-\cos \left(\frac{x^2}{2}ight)-\cos \left(\frac{x^2}{4}ight)+\cos \left(\frac{x^2}{2}ight) \cdot \cos \left(\frac{x^2}{4}ight)ight]$કૌંસમાં રહેલા પદોનું અવયવીકરણ કરતા:$L = \lim _{x}$ ${ightarrow 0} \frac{8}{x^8}\left[\left(1-\cos \frac{x^2}{2}ight) - \cos \left(\frac{x^2}{4}ight)\left(1-\cos \frac{x^2}{2}ight)ight]$$L = \lim _{x ightarrow 0} \frac{8}{x^8}\left(1-\cos \frac{x^2}{2}ight)\left(1-\cos \frac{x^2}{4}ight)$નિત્યસમ $1-\cos 	heta = 2 \sin^2 \left(\frac{	heta}{2}ight)$ નો ઉપયોગ કરતા:$L = \lim _{x ightarrow 0} \frac{8}{x^8} \left(2 \sin^2 \frac{x^2}{4}ight) \left(2 \sin^2 \frac{x^2}{8}ight)$$L = \lim _{x ightarrow 0} \frac{32}{x^8} \left(\sin^2 \frac{x^2}{4}ight) \left(\sin^2 \frac{x^2}{8}ight)$$\left(\frac{x^2}{4}ight)^2$ અને $\left(\frac{x^2}{8}ight)^2$ વડે ગુણતા અને ભાગતા:$L = 32 \lim _{x}$ ${ightarrow 0} \left(\frac{\sin \frac{x^2}{4}}{\frac{x^2}{4}}ight)^2 \left(\frac{x^4}{16}ight) \left(\frac{\sin \frac{x^2}{8}}{\frac{x^2}{8}}ight)^2 \left(\frac{x^4}{64}ight) \cdot \frac{1}{x^8}$$L = 32 \cdot (1)^2 \cdot \frac{1}{16} \cdot (1)^2 \cdot \frac{1}{64}$$L = 32 \cdot \frac{1}{1024} = \frac{1}{32}$

$\lim _{x}$ ${\rightarrow 0} \frac{8}{x^8}\left[1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^2}{4}\right)\right]$ ની કિંમત શોધો.

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