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

Let f(x) = (1-x(1+|1-x|))/(|1-x|)cos(1/(1-x)) for x ≠ 1. Then $\text{Given: } f(x) = \dfrac{1-x(1+|1-x|)}{|1-x|}\cos\!\left(\dfrac{1}{1-x}\right),\quad x \neq 1$ $\text{study one-sided limits as } x \to 1^- \text{ and } x \to 1^+$ $\text{To find: which of (1)--(4) are TRUE}$ $1.\ \text{simplify with } h = x-1$ $2.\ \text{left limit } (x \to 1^-)$ $3.\ \text{right limit } (x \to 1^+),\ \text{then decide}$ $h = x-1,\quad |1-x| = |h|,\quad \dfrac{1}{1-x} = -\dfrac{1}{h}$ $f = \dfrac{1-(1+h)(1+|h|)}{|h|}\cos\!\left(-\dfrac{1}{h}\right)$ $1 - (1+h)(1+|h|) = -h - (1+h)|h|$ $\dfrac{1-x(1+|1-x|)}{|1-x|} = \dfrac{-h-(1+h)|h|}{|h|}$ $= -\dfrac{h}{|h|} - (1+h)$ $x \to 1^-:\ h < 0,\ |h| = -h \Rightarrow \dfrac{h}{|h|} = -1$ $\dfrac{-h-(1+h)|h|}{|h|} = 1-(1+h) = -h \to 0$ $\left|\cos\!\left(-\dfrac{1}{h}\right)\right| \le 1$ $-h \to 0 \Rightarrow \lim_{x\to1^-} f(x) = 0$ $\lim_{x\to1^-} f(x) = 0 \Rightarrow \text{(1) TRUE},\ \text{(2) FALSE}$ $x \to 1^+:\ h > 0,\ |h| = h \Rightarrow \dfrac{h}{|h|} = 1$ $\dfrac{-h-(1+h)|h|}{|h|} = -1-(1+h) = -(2+h) \to -2$ $f \approx -2\cos\!\left(\dfrac{1}{h}\right) \text{ as } h \to 0^+$ $\cos(1/h) \text{ oscillates} \Rightarrow \lim_{x\to1^+} f(x) \text{ does not exist}$ $\lim_{x\to1^+} f(x) \text{ does not exist} \Rightarrow \text{(4) TRUE},\ \text{(3) FALSE}$ $\boxed{\text{Answer: (1) and (4)}}$