If {x} = x - [x] where [x] is the greatest in | vedclass

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(A) Let $h = -x$,where $h > 0$ as $x \rightarrow 0^-$. Since $x$ is slightly less than $0$,$[x] = -1$. Thus,$\{x\} = x - (-1) = x + 1 = 1 - h$. As $x

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$\frac{1}{\sqrt{3}}$

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(A) Let $h = -x$,where $h > 0$ as $x ightarrow 0^-$. Since $x$ is slightly less than $0$,$[x] = -1$. Thus,$\{x\} = x - (-1) = x + 1 = 1 - h$. As $x ightarrow 0^-$,$h ightarrow 0^+$,so $\{x\} ightarrow 1^-$. Let $u = \{x\}$. As $u ightarrow 1^-$,the limit becomes $\lim_{u ightarrow 1^-} \frac{\cos^{-1}(1-u^2) \sin^{-1}(1-u)}{u(1-u^3)} = \lim_{u ightarrow 1^-} \frac{\cos^{-1}(1-u^2) \sin^{-1}(1-u)}{u(1-u)(1+u+u^2)}$. Let $t = 1-u$. As $u ightarrow 1$,$t ightarrow 0^+$. Then $1-u^2 = (1-u)(1+u) = t(2-t)$. As $t ightarrow 0^+$,$\cos^{-1}(1-u^2) = \cos^{-1}(t(2-t)) ightarrow \cos^{-1}(0) = \frac{\pi}{2}$. Also,$\sin^{-1}(1-u) = \sin^{-1}(t) \approx t$ as $t ightarrow 0$. Substituting these into the limit: $\lim_{t ightarrow 0^+} \frac{(\frac{\pi}{2}) \cdot t}{(1-t)(t)(1+(1-t)+(1-t)^2)} = \frac{\pi/2}{1 \cdot 3} = \frac{\pi}{6}$. Thus,$	heta = \frac{\pi}{6}$. Therefore,$	an 	heta = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.

If $\{x\} = x - [x]$ where $[x]$ is the greatest integer $\leq x$ and $\lim_{x \rightarrow 0^{-}} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^4} = \theta$,then $\tan \theta =$

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