If lim _{x rightarrow 0} frac{2 a sin x-sin 2 | vedclass

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(B) Given the limit: $\lim _{x ightarrow 0} \frac{2 a \sin x-\sin 2 x}{	an ^{3} x} = 1$. Using the Taylor series expansion for $\sin x = x - \frac{

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$1$

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(B) Given the limit: $\lim _{x ightarrow 0} \frac{2 a \sin x-\sin 2 x}{	an ^{3} x} = 1$. Using the Taylor series expansion for $\sin x = x - \frac{x^3}{6} + \dots$ and $\sin 2x = 2x - \frac{8x^3}{6} + \dots$ and $	an x = x + \frac{x^3}{3} + \dots$: $\lim _{x ightarrow 0} \frac{2 a(x - \frac{x^3}{6}) - (2x - \frac{8x^3}{6})}{x^3} = 1$. $\lim _{x ightarrow 0} \frac{(2a - 2)x + (\frac{8}{6} - \frac{2a}{6})x^3}{x^3} = 1$. For the limit to exist, the coefficient of $x$ must be $0$, so $2a - 2 = 0$, which gives $a = 1$. Substituting $a = 1$ into the expression: $\lim _{x ightarrow 0} \frac{(8/6 - 2/6)x^3}{x^3} = \frac{6/6}{1} = 1$. Thus, the value of $a$ is $1$.

If $\lim _{x \rightarrow 0} \frac{2 a \sin x-\sin 2 x}{\tan ^{3} x}$ exists and is equal to $1$, then the value of $a$ is

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