Let f(x) = frac{1 - x(1 + |1 - x|)}{|1 - x|} | vedclass

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(B) For $x 	o 1^{-}$,let $x = 1 - h$ where $h > 0$. Then $|1 - x| = h$. $f(1 - h) = \frac{1 - (1 - h)(1 + h)}{h} \cos \left(\frac{1}{h}ight) = \fra

Vedclass · Accepted Answer

$A, D$

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(B) For $x 	o 1^{-}$,let $x = 1 - h$ where $h > 0$. Then $|1 - x| = h$. $f(1 - h) = \frac{1 - (1 - h)(1 + h)}{h} \cos \left(\frac{1}{h}ight) = \frac{1 - (1 - h^2)}{h} \cos \left(\frac{1}{h}ight) = \frac{h^2}{h} \cos \left(\frac{1}{h}ight) = h \cos \left(\frac{1}{h}ight)$. Since $\lim_{h 	o 0} h \cos \left(\frac{1}{h}ight) = 0$,we have $\lim_{x 	o 1^{-}} f(x) = 0$. For $x 	o 1^{+}$,let $x = 1 + h$ where $h > 0$. Then $|1 - x| = h$. $f(1 + h) = \frac{1 - (1 + h)(1 + h)}{h} \cos \left(\frac{-1}{h}ight) = \frac{1 - (1 + 2h + h^2)}{h} \cos \left(\frac{1}{h}ight) = \frac{-2h - h^2}{h} \cos \left(\frac{1}{h}ight) = -(2 + h) \cos \left(\frac{1}{h}ight)$. As $h 	o 0$,$-(2 + h) 	o -2$,and $\cos \left(\frac{1}{h}ight)$ oscillates between $-1$ and $1$. Thus,$\lim_{x 	o 1^{+}} f(x)$ does not exist. Therefore,options $A$ and $D$ are correct.

Let $f(x) = \frac{1 - x(1 + |1 - x|)}{|1 - x|} \cos \left(\frac{1}{1 - x}\right)$ for $x \neq 1$. Then

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