The value of lim _{x rightarrow 0} frac{(1-co | vedclass

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(A) We need to evaluate the limit: $\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \sin 5 x}{x^2 \sin 3 x}$.Using the identity $1-\cos 2x = 2\sin^2 x$,we

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$\frac{10}{3}$

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(A) We need to evaluate the limit: $\lim _{x ightarrow 0} \frac{(1-\cos 2 x) \sin 5 x}{x^2 \sin 3 x}$.Using the identity $1-\cos 2x = 2\sin^2 x$,we rewrite the expression:$= \lim _{x ightarrow 0} \frac{2 \sin^2 x \cdot \sin 5 x}{x^2 \sin 3 x}$.$= 2 \cdot \lim _{x ightarrow 0} \left( \frac{\sin x}{x} ight)^2 \cdot \frac{\sin 5 x}{x} \cdot \frac{x}{\sin 3 x}$.Multiply and divide by $5$ and $3$ to use the standard limit $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$:$= 2 \cdot \left( \lim _{x ightarrow 0} \frac{\sin x}{x} ight)^2 \cdot 5 \left( \lim _{x ightarrow 0} \frac{\sin 5 x}{5 x} ight) \cdot \frac{1}{3} \left( \lim _{x ightarrow 0} \frac{3 x}{\sin 3 x} ight)$.$= 2 \cdot (1)^2 \cdot 5 \cdot (1) \cdot \frac{1}{3} \cdot (1) = \frac{10}{3}$.

The value of $\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \sin 5 x}{x^2 \sin 3 x}$ is

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