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

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(B) We have the limit $L = \lim _{x ightarrow 0} \frac{	an ^3 x-\sin ^3 x}{x^5}$.Using the Taylor series expansions: $	an x = x + \frac{x^3}{3} +

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$\frac{3}{2}$

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(B) We have the limit $L = \lim _{x ightarrow 0} \frac{	an ^3 x-\sin ^3 x}{x^5}$.Using the Taylor series expansions: $	an x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)$ and $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7)$.Factorizing the numerator: $	an^3 x - \sin^3 x = (	an x - \sin x)(	an^2 x + 	an x \sin x + \sin^2 x)$.Now,$	an x - \sin x = \sin x (\sec x - 1) = \sin x \left(\frac{1-\cos x}{\cos x}ight) = \sin x \cdot \frac{2 \sin^2(x/2)}{\cos x}$.As $x ightarrow 0$,$\sin x \approx x$,$\sin(x/2) \approx x/2$,and $\cos x \approx 1$.So,$	an x - \sin x \approx x \cdot \frac{2(x/2)^2}{1} = \frac{x^3}{2}$.Substituting this into the limit expression:$L = \lim _{x ightarrow 0} \frac{(\frac{x^3}{2})(	an^2 x + 	an x \sin x + \sin^2 x)}{x^5}$.$L = \lim _{x ightarrow 0} \frac{1}{2} \cdot \frac{	an^2 x + 	an x \sin x + \sin^2 x}{x^2}$.Since $\lim _{x ightarrow 0} \frac{	an x}{x} = 1$ and $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$,we have:$L = \frac{1}{2} (1^2 + 1 \cdot 1 + 1^2) = \frac{1}{2} (3) = \frac{3}{2}$.

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

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