lim _{x rightarrow 0} frac{cos 4 x-4 cos 2 x+ | vedclass

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Question

(B) Given the limit: $\lim _{x \rightarrow 0} \frac{\cos 4 x-4 \cos 2 x+3}{x^4}$ which is in the $\frac{0}{0}$ indeterminate form. Using Taylor series

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Given the limit: $\lim _{x ightarrow 0} \frac{\cos 4 x-4 \cos 2 x+3}{x^4}$ which is in the $\frac{0}{0}$ indeterminate form. Using Taylor series expansion for $\cos 	heta \approx 1 - \frac{	heta^2}{2!} + \frac{	heta^4}{4!}$: $\cos 4x \approx 1 - \frac{(4x)^2}{2} + \frac{(4x)^4}{24} = 1 - 8x^2 + \frac{256x^4}{24} = 1 - 8x^2 + \frac{32x^4}{3}$. $\cos 2x \approx 1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24} = 1 - 2x^2 + \frac{16x^4}{24} = 1 - 2x^2 + \frac{2x^4}{3}$. Substituting these into the numerator: $(1 - 8x^2 + \frac{32x^4}{3}) - 4(1 - 2x^2 + \frac{2x^4}{3}) + 3$ $= 1 - 8x^2 + \frac{32x^4}{3} - 4 + 8x^2 - \frac{8x^4}{3} + 3$ $= (1 - 4 + 3) + (-8x^2 + 8x^2) + (\frac{32x^4}{3} - \frac{8x^4}{3})$ $= 0 + 0 + \frac{24x^4}{3} = 8x^4$. Thus,$\lim _{x ightarrow 0} \frac{8x^4}{x^4} = 8$.

$\lim _{x \rightarrow 0} \frac{\cos 4 x-4 \cos 2 x+3}{x^4} = $

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