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

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Question

(B) We know that $x^\circ = \frac{\pi x}{180} 	ext{ radians}$.Therefore,$\mathop {\lim }\limits_{x 	o 0} \frac{{\sin {x^\circ}}}{x} = \mathop {\lim

Vedclass · Accepted Answer

$\frac{\pi}{180}$

Vedclass · Answer

(B) We know that $x^\circ = \frac{\pi x}{180} 	ext{ radians}$.Therefore,$\mathop {\lim }\limits_{x 	o 0} \frac{{\sin {x^\circ}}}{x} = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin \left( \frac{\pi x}{180} ight)}}{x}$.Multiplying and dividing by $\frac{\pi}{180}$,we get:$\mathop {\lim }\limits_{x 	o 0} \left( \frac{\pi}{180} ight) \cdot \frac{{\sin \left( \frac{\pi x}{180} ight)}}{\frac{\pi x}{180}}$.Since $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$,the expression becomes $\frac{\pi}{180} \cdot 1 = \frac{\pi}{180}$.

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

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