We need to study the one-sided limits of

$$f(x)=\frac{1-x(1+|1-x|)}{|1-x|}\cos\left(\frac{1}{1-x}\right), \qquad x\ne 1.$$

We will simplify separately for $x<1$ and $x>1$.

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1. Case $x\to 1^-$

If $x<1$, then $|1-x|=1-x.$ So,

$$1-x(1+|1-x|)=1-x\bigl(1+(1-x)\bigr)=1-x(2-x).$$

Now simplify:

$$1-x(2-x)=1-2x+x^2=(1-x)^2.$$

Therefore,

$$\frac{1-x(1+|1-x|)}{|1-x|}=\frac{(1-x)^2}{1-x}=1-x.$$

Hence for $x<1$,

$$f(x)=(1-x)\cos\left(\frac{1}{1-x}\right).$$

Now as $x\to 1^-$, $1-x\to 0^+,$ and we know

$$-1\le \cos\left(\frac{1}{1-x}\right)\le 1.$$

So,

$$-(1-x)\le f(x)\le (1-x).$$

By squeeze theorem,

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

So:

• B is false
• C is true

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2. Case $x\to 1^+$

If $x>1$, then $|1-x|=x-1.$ So,

$$1-x(1+|1-x|)=1-x\bigl(1+(x-1)\bigr)=1-x\cdot x=1-x^2.$$

Thus,

$$\frac{1-x(1+|1-x|)}{|1-x|}=\frac{1-x^2}{x-1}.$$

Factor numerator:

$$1-x^2=-(x^2-1)=-(x-1)(x+1).$$

Hence,

$$\frac{1-x^2}{x-1}=-(x+1).$$

Therefore for $x>1$,

$$f(x)=-(x+1)\cos\left(\frac{1}{1-x}\right).$$

As $x\to 1^+$, $x+1\to 2,$ so the prefactor tends to $-2$, not to $0$. Meanwhile,

$$\cos\left(\frac{1}{1-x}\right)$$

oscillates because $\frac{1}{1-x}\to -\infty$ as $x\to 1^+$.

To show the limit does not exist, take two sequences approaching $1^+$.

Sequence 1

Let

$$\frac{1}{1-x_n}=-2n\pi.$$

Then $x_n\to 1^+$ and

$$\cos\left(\frac{1}{1-x_n}\right)=\cos(-2n\pi)=1.$$

So,

$$f(x_n)=-(x_n+1)\to -2.$$

Sequence 2

Let

$$\frac{1}{1-y_n}=-(2n+1)\pi.$$

Then $y_n\to 1^+$ and

$$\cos\left(\frac{1}{1-y_n}\right)=\cos(-(2n+1)\pi)=-1.$$

So,

$$f(y_n)=-(y_n+1)(-1)=y_n+1\to 2.$$

Since two different sequences approaching $1^+$ give different limit values ($-2$ and $2$),

$$\lim_{x\to 1^+} f(x)$$

does not exist.

So:

• A is false
• D is true

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3. Final conclusion

The correct options are: $\boxed{C, D}$

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4. Comparison with stored answer

Stored correct answer: C, D

Our derived answer: C, D

So the derived answer agrees with the stored answer.