For each t in mathbb{R},let [t] be the greate | vedclass

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(B) We are evaluating the limit $\lim_{x 	o 1^+} \frac{(1 - |x| + \sin |1 - x|) \sin (\frac{\pi}{2} [1 - x])}{|1 - x|^2}$.As $x 	o 1^+$,we have $x >

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equals $0$

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(B) We are evaluating the limit $\lim_{x 	o 1^+} \frac{(1 - |x| + \sin |1 - x|) \sin (\frac{\pi}{2} [1 - x])}{|1 - x|^2}$.As $x 	o 1^+$,we have $x > 1$,so $|x| = x$ and $|1 - x| = x - 1$.Also,for $x$ slightly greater than $1$,$1 - x$ is slightly less than $0$,so $[1 - x] = -1$.Substituting these into the expression:$\lim_{x 	o 1^+} \frac{(1 - x + \sin(x - 1)) \sin(-\frac{\pi}{2})}{(x - 1)^2}$Since $\sin(-\frac{\pi}{2}) = -1$,the expression becomes:$\lim_{x 	o 1^+} \frac{-(1 - x + \sin(x - 1))}{(x - 1)^2} = \lim_{x 	o 1^+} \frac{(x - 1) - \sin(x - 1)}{(x - 1)^2}$Let $h = x - 1$. As $x 	o 1^+$,$h 	o 0^+$. The limit becomes:$\lim_{h 	o 0^+} \frac{h - \sin(h)}{h^2}$Using the Taylor series expansion $\sin(h) = h - \frac{h^3}{6} + \dots$:$\lim_{h 	o 0^+} \frac{h - (h - \frac{h^3}{6} + \dots)}{h^2} = \lim_{h 	o 0^+} \frac{\frac{h^3}{6}}{h^2} = \lim_{h 	o 0^+} \frac{h}{6} = 0$.

For each $t \in \mathbb{R}$,let $[t]$ be the greatest integer less than or equal to $t$. Then $\lim_{x \to 1^+} \frac{(1 - |x| + \sin |1 - x|) \sin (\frac{\pi}{2} [1 - x])}{|1 - x|^2}$ is:

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