If the function f(x) = frac{tan(tan x) - sin( | vedclass

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(B) For $f(x)$ to be continuous at $x = 0$,$f(0) = \lim_{x 	o 0} f(x)$.We know the expansions: $	an x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$

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$2$

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(B) For $f(x)$ to be continuous at $x = 0$,$f(0) = \lim_{x 	o 0} f(x)$.We know the expansions: $	an x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$ and $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$Numerator: $	an(	an x) - \sin(\sin x) = (	an x + \frac{	an^3 x}{3} + \dots) - (\sin x - \frac{\sin^3 x}{6} + \dots) = (	an x - \sin x) + \frac{	an^3 x}{3} + \frac{\sin^3 x}{6} + \dots$Denominator: $	an x - \sin x = (x + \frac{x^3}{3} + \dots) - (x - \frac{x^3}{6} + \dots) = \frac{x^3}{2} + \dots$Thus,$f(x) = \frac{(	an x - \sin x) + \frac{	an^3 x}{3} + \frac{\sin^3 x}{6}}{	an x - \sin x} = 1 + \frac{\frac{	an^3 x}{3} + \frac{\sin^3 x}{6}}{	an x - \sin x}$.Dividing numerator and denominator by $x^3$: $f(x) = 1 + \frac{\frac{1}{3}(\frac{	an x}{x})^3 + \frac{1}{6}(\frac{\sin x}{x})^3}{\frac{	an x - \sin x}{x^3}} = 1 + \frac{\frac{1}{3} + \frac{1}{6}}{\frac{1}{2}} = 1 + \frac{1/2}{1/2} = 1 + 1 = 2$.

If the function $f(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}$ is continuous at $x = 0$,then $f(0)$ is equal to . . . . . . .

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