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(B) Let $L = \lim _{x \rightarrow 0} \left( \frac{1-\cos ^2(3 x)}{\cos ^3(4 x)} \cdot \frac{\sin ^3(4 x)}{(\ln(1+2 x))^5} \right)$.Using the identity

Vedclass · Accepted Answer

$18$

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(B) Let $L = \lim _{x ightarrow 0} \left( \frac{1-\cos ^2(3 x)}{\cos ^3(4 x)} \cdot \frac{\sin ^3(4 x)}{(\ln(1+2 x))^5} ight)$.Using the identity $1-\cos^2 	heta = \sin^2 	heta$,we have:$L = \lim _{x ightarrow 0} \frac{\sin^2(3x)}{\cos^3(4x)} \cdot \frac{\sin^3(4x)}{(\ln(1+2x))^5}$.Multiply and divide by the necessary terms to use standard limits $\lim_{u 	o 0} \frac{\sin u}{u} = 1$ and $\lim_{u 	o 0} \frac{\ln(1+u)}{u} = 1$:$L = \lim _{x}$ ${ightarrow 0} \left( \frac{\sin^2(3x)}{(3x)^2} \cdot (3x)^2 ight) \cdot \frac{1}{\cos^3(4x)} \cdot \left( \frac{\sin^3(4x)}{(4x)^3} \cdot (4x)^3 ight) \cdot \left( \frac{2x}{\ln(1+2x)} ight)^5 \cdot \frac{1}{(2x)^5}$.$L = \lim _{x}$ ${ightarrow 0} \left( 1^2 \cdot 9x^2 ight) \cdot \frac{1}{1} \cdot \left( 1^3 \cdot 64x^3 ight) \cdot 1^5 \cdot \frac{1}{32x^5}$.$L = \lim _{x ightarrow 0} \frac{9 \cdot 64 \cdot x^5}{32 \cdot x^5} = \frac{576}{32} = 18$.

$\lim _{x}$ ${\rightarrow 0} \left( \left( \frac{1-\cos ^2(3 x)}{\cos ^3(4 x)} \right) \left( \frac{\sin ^3(4 x)}{(\log _e(2 x+1))^5} \right) \right)$ is equal to $.........$.

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