The value of mathop {lim }limits_{x to 0} fra | vedclass

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(A) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{(1 - \cos 2x)\sin 5x}{x^2 \sin 3x}$ Using the identity $1 - \cos 2x = 2 \sin^2 x$,we get: $=

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$10/3$

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(A) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{(1 - \cos 2x)\sin 5x}{x^2 \sin 3x}$ Using the identity $1 - \cos 2x = 2 \sin^2 x$,we get: $= \mathop {\lim }\limits_{x 	o 0} \frac{2 \sin^2 x \sin 5x}{x^2 \sin 3x}$ $= \mathop {\lim }\limits_{x 	o 0} 2 \left( \frac{\sin x}{x} ight)^2 \cdot \frac{\sin 5x}{\sin 3x}$ Multiply and divide by $5x$ and $3x$: $= \mathop {\lim }\limits_{x 	o 0} 2 \left( \frac{\sin x}{x} ight)^2 \cdot \frac{\frac{\sin 5x}{5x} \cdot 5x}{\frac{\sin 3x}{3x} \cdot 3x}$ Since $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$: $= 2 \cdot (1)^2 \cdot \frac{1 \cdot 5}{1 \cdot 3} = \frac{10}{3}$.

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

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