mathop {lim }limits_{x to 0} frac{{3sin x - s | vedclass

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Question

(A) We know the trigonometric identity $\sin 3x = 3\sin x - 4\sin^3 x$. Substituting this into the expression: $\mathop {\lim }\limits_{x 	o 0} \frac

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) We know the trigonometric identity $\sin 3x = 3\sin x - 4\sin^3 x$. Substituting this into the expression: $\mathop {\lim }\limits_{x 	o 0} \frac{{3\sin x - (3\sin x - 4\sin^3 x)}}{{{x^3}}}$ $= \mathop {\lim }\limits_{x 	o 0} \frac{{4\sin^3 x}}{{{x^3}}}$ $= 4 	imes \mathop {\lim }\limits_{x 	o 0} \left( \frac{\sin x}{x} ight)^3$ Since $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$,we get: $= 4 	imes (1)^3 = 4$.

$\mathop {\lim }\limits_{x \to 0} \frac{{3\sin x - \sin 3x}}{{{x^3}}} = $

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