Let f x = 1 - x ( 1 + 1 - x ) | 1 - x | cos ⁡ 1 1 - x   for   x ≠ 1 , then

$f\left(x\right)= \left\{\begin{array}{cc}\left(1-x\right)\cos \frac{1}{1-x},& x<1\\ -\left(1+x\right)\cos \frac{1}{1-x},& x>1\end{array}\right.$

$\underset{x\to 1^{+}}{\lim }f\left(x\right)=\mathrm{does\; not\; exist}$; (as $\text{cos}\text{t}$ can take any value between -1 and 1, so oscillating.)

$\underset{x\to 1^{-}}{\lim }f\left(x\right)=0$: (As $\text{cos}\text{t}$ term varies between -1 and 1 for any value of t and is multiplied with zero)