Let alpha in R be such that the function f(x) | vedclass

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(C) For the function to be continuous at $x=0$,we must have $\operatorname{Lim}_{x \rightarrow 0^{+}} f(x) = \operatorname{Lim}_{x \rightarrow 0^{-}}

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no such $\alpha$ exists

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(C) For the function to be continuous at $x=0$,we must have $\operatorname{Lim}_{x ightarrow 0^{+}} f(x) = \operatorname{Lim}_{x ightarrow 0^{-}} f(x) = f(0) = \alpha$.First,consider the Right Hand Limit $(RHL)$ as $x ightarrow 0^{+}$:$\operatorname{Lim}_{x ightarrow 0^{+}} \frac{\cos^{-1}(1-x^2) \sin^{-1}(1-x)}{x(1-x)(1+x)} = \operatorname{Lim}_{x ightarrow 0^{+}} \frac{\cos^{-1}(1-x^2)}{x} \cdot \frac{\sin^{-1}(1-x)}{1-x^2} = \frac{\pi}{2} \cdot \operatorname{Lim}_{x ightarrow 0^{+}} \frac{\cos^{-1}(1-x^2)}{x}$.Let $1-x^2 = \cos 	heta$,then as $x ightarrow 0^{+}$,$	heta ightarrow 0^{+}$.$\operatorname{Lim}_{	heta ightarrow 0^{+}} \frac{	heta}{\sqrt{1-\cos 	heta}} = \operatorname{Lim}_{	heta ightarrow 0^{+}} \frac{	heta}{\sqrt{2} \sin(	heta/2)} = \frac{1}{\sqrt{2}} \cdot 2 = \sqrt{2}$.So,$RHL = \frac{\pi}{2} \cdot \sqrt{2} = \frac{\pi}{\sqrt{2}}$.Now,consider the Left Hand Limit $(LHL)$ as $x ightarrow 0^{-}$:For $x \in (-1, 0)$,$\{x\} = x+1$.$\operatorname{Lim}_{x ightarrow 0^{-}} \frac{\cos^{-1}(1-(x+1)^2) \sin^{-1}(1-(x+1))}{(x+1)-(x+1)^3} = \operatorname{Lim}_{x ightarrow 0^{-}} \frac{\cos^{-1}(1-(x+1)^2) \sin^{-1}(-x)}{(x+1)(1-(x+1)^2)} = \operatorname{Lim}_{x ightarrow 0^{-}} \frac{\cos^{-1}(1-(x+1)^2) \cdot (-x)}{(x+1)(-x)(2+x)} = \frac{\cos^{-1}(0)}{1 \cdot 2} = \frac{\pi/2}{2} = \frac{\pi}{4}$.Since $RHL = \frac{\pi}{\sqrt{2}}$ and $LHL = \frac{\pi}{4}$,the limits are not equal.Therefore,the function is not continuous at $x=0$ for any value of $\alpha$.

Let $\alpha \in R$ be such that the function $f(x) = \begin{cases} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, & x \neq 0 \\ \alpha, & x=0 \end{cases}$ is continuous at $x=0$,where $\{x\} = x - [x]$ and $[x]$ is the greatest integer less than or equal to $x$. Then:

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