If f(x)= begin{cases} frac{x-[x]}{x-2}, & x2 | vedclass

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(C) Given $f(x)$ is continuous on $R$. For $x=2$,$\lim _{x \rightarrow 2^{+}} f(x) = f(2)$.$\lim _{x \rightarrow 2^{+}} \frac{x-[x]}{x-2}$. Since $x \

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$2$

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(C) Given $f(x)$ is continuous on $R$. For $x=2$,$\lim _{x ightarrow 2^{+}} f(x) = f(2)$.$\lim _{x ightarrow 2^{+}} \frac{x-[x]}{x-2}$. Since $x ightarrow 2^{+}$,$[x]=2$,so $\lim _{x ightarrow 2^{+}} \frac{x-2}{x-2} = 1$. Thus,$b=1$.For $x=2$,$\lim _{x ightarrow 2^{-}} f(x) = f(2)$.$\lim _{x ightarrow 2^{-}} \frac{|(x-2)(x+1)|}{a(2+x-x^2)} = \lim _{x ightarrow 2^{-}} \frac{|(x-2)(x+1)|}{-a(x-2)(x+1)}$.Since $x ightarrow 2^{-}$,$x-2 $\lim _{x ightarrow 2^{-}} \frac{-(x-2)(x+1)}{-a(x-2)(x+1)} = \frac{1}{a} = 1 \Rightarrow a=1$.Now,we evaluate $\lim _{x ightarrow 0} \frac{\sin ^2 ax+x 	an bx}{x^2}$ with $a=1, b=1$.$\lim _{x ightarrow 0} \frac{\sin ^2 x+x 	an x}{x^2} = \lim _{x ightarrow 0} \left( \frac{\sin ^2 x}{x^2} + \frac{	an x}{x} ight) = 1^2 + 1 = 2$.

If $f(x)= \begin{cases} \frac{x-[x]}{x-2}, & x>2 \\ b, & x=2 \\ \frac{|x^2-x-2|}{a(2+x-x^2)}, & -1 < x \leq 2 \\ 2a-b, & x \leq -1 \end{cases}$ is continuous on $R$,then $\lim _{x \rightarrow 0} \frac{\sin ^2 ax+x \tan bx}{x^2}=$

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