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Let

$f(x)= \frac{1-x(1+\mid1-x\mid)}{\mid1-x\mid}\cos(\frac{1}{1-x})$ for x≠ 1, then

$\lim_{x \rightarrow 1^{+}}f(x)=0$

$\lim_{x \rightarrow 1^{+}}f(x)$ doesnot exist

$\lim_{x \rightarrow 1^{-}}f(x)=0$

$\lim_{x \rightarrow 1^{-}}f(x)$ doesnot exist

Option B,C

$f(x)= \frac{1-x(1+\mid1-x\mid)}{\mid1-x\mid}\cos(\frac{1}{1-x})$

Now,

$\lim_{x \rightarrow 1^{-}}f(x)$

$= \lim_{x \rightarrow 1^{-}}\frac{1-x(1+1-x)}{\mid1-x\mid}\cos(\frac{1}{1-x})$

$= \lim_{x \rightarrow 1^{-}}(1-x)\cos(\frac{1}{1-x})=0 and \lim_{x \rightarrow 1^{+}}f(x)=\lim_{x \rightarrow 1^{+}}$

$\frac{1-x(1-1+x)}{x-1}\cos(\frac{1}{1-x})$

$=\lim_{x \rightarrow 1^{+}}-(x+1) cos (\frac{1}{x+1})$ where x≠ 1 doesnot exist

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