lim _{x rightarrow 0} frac{tan ^3 x - sin ^3 | vedclass

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(B) We use the Taylor series expansions for $	an x$ and $\sin x$ near $x = 0$: $	an x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)$ $\sin x = x -

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$\frac{3}{2}$

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(B) We use the Taylor series expansions for $	an x$ and $\sin x$ near $x = 0$: $	an x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)$ $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7)$ Now,expand $	an^3 x$: $	an^3 x = (x + \frac{x^3}{3} + O(x^5))^3 = x^3 + 3x^2(\frac{x^3}{3}) + O(x^7) = x^3 + x^5 + O(x^7)$ Next,expand $\sin^3 x$: $\sin^3 x = (x - \frac{x^3}{6} + O(x^5))^3 = x^3 + 3x^2(-\frac{x^3}{6}) + O(x^7) = x^3 - \frac{x^5}{2} + O(x^7)$ Subtracting these two expansions: $	an^3 x - \sin^3 x = (x^3 + x^5) - (x^3 - \frac{x^5}{2}) + O(x^7) = \frac{3}{2}x^5 + O(x^7)$ Finally,evaluate the limit: $\lim _{x ightarrow 0} \frac{\frac{3}{2}x^5}{x^5} = \frac{3}{2}$.

$\lim _{x \rightarrow 0} \frac{\tan ^3 x - \sin ^3 x}{x^5}$ is equal to

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