We need to analyze the given function $f(x)$ and determine the behavior of its limit as $x \to 1^+$ and $x \to 1^-$. The function is given as:

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

Step 1: Analyze the expression for $f(x)$

1. Break it into parts:

   • The numerator is $1 - x (1 + |1 - x|)$.
   • The denominator is $|1 - x|$.
   • The cosine term $\cos\left(\frac{1}{1 - x}\right)$ oscillates between $-1$ and $1$.

2. Behavior of $|1 - x|$ as $x \to 1^+$ and $x \to 1^-$:

   • For $x \to 1^+$, $|1 - x| = x - 1$ (since $1 - x < 0$).
   • For $x \to 1^-$, $|1 - x| = 1 - x$ (since $1 - x > 0$).

Step 2: Behavior of $f(x)$ as $x \to 1^+$

For $x \to 1^+$:

• The numerator becomes $1 - x (1 + (x - 1)) = 1 - x (x) = 1 - x^2$.
• The denominator is $x - 1$.
• The function becomes: $f(x) = \frac{1 - x^2}{x - 1} \cos\left(\frac{1}{1 - x}\right).$

Simplify $\frac{1 - x^2}{x - 1}$: $1 - x^2 = (1 - x)(1 + x), \quad \therefore \frac{1 - x^2}{x - 1} = -(1 + x).$

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

As $x \to 1^+$:

• $1 + x \to 2$, but $\cos\left(\frac{1}{1 - x}\right)$ oscillates between $-1$ and $1$.
• Hence, $f(x)$ oscillates between $-2$ and $2$.

Conclusion: The limit $\lim_{x \to 1^+} f(x)$ does not exist.

Step 3: Behavior of $f(x)$ as $x \to 1^-$

For $x \to 1^-$:

• The numerator becomes $1 - x (1 + (1 - x)) = 1 - x (2 - x) = 1 - 2x + x^2$.
• The denominator is $1 - x$.
• The function becomes: $f(x) = \frac{1 - 2x + x^2}{1 - x} \cos\left(\frac{1}{1 - x}\right).$

Simplify $\frac{1 - 2x + x^2}{1 - x}$: $1 - 2x + x^2 = (1 - x)^2, \quad \therefore \frac{1 - 2x + x^2}{1 - x} = 1 - x.$

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

As $x \to 1^-$:

• $1 - x \to 0$, but $\cos\left(\frac{1}{1 - x}\right)$ oscillates between $-1$ and $1$.
• Hence, $f(x) \to 0$, but the oscillations vanish due to the factor $1 - x$.

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

Step 4: Final Answer

• $\lim_{x \to 1^+} f(x)$ does not exist.
• $\lim_{x \to 1^-} f(x) = 0$.

Thus, the correct option is (2) $\lim_{x \to 1^+} f(x)$ does not exist.

Would you like further clarification or a detailed plot of this behavior? Here are some related questions to explore:

1. Why does the oscillating cosine term cause the limit to fail for $x \to 1^+$?
2. How would the function change if $\cos\left(\frac{1}{1 - x}\right)$ were replaced with a bounded function?
3. What role does $|1 - x|$ play in ensuring continuity for $x \neq 1$?
4. Can you derive similar limits for functions involving sine or other trigonometric oscillations?
5. What happens to the behavior if we define $f(1)$ explicitly?

Tip: Always check for oscillating terms like sine or cosine when evaluating limits—they can lead to non-existence if unbounded or fast-oscillating.