mathop {lim }limits_{x to 0} frac{sin 2x}{x} | vedclass

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(D) We know that $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$. To evaluate $\mathop {\lim }\limits_{x 	o 0} \frac{\sin 2x}{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) We know that $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$. To evaluate $\mathop {\lim }\limits_{x 	o 0} \frac{\sin 2x}{x}$,we multiply and divide by $2$: $\mathop {\lim }\limits_{x 	o 0} \frac{\sin 2x}{x} = \mathop {\lim }\limits_{x 	o 0} \left( 2 	imes \frac{\sin 2x}{2x} ight)$. Since as $x 	o 0$,$2x 	o 0$,we have: $2 	imes \mathop {\lim }\limits_{2x 	o 0} \frac{\sin 2x}{2x} = 2 	imes 1 = 2$.

$\mathop {\lim }\limits_{x \to 0} \frac{\sin 2x}{x} = $

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