Let `f(x) = ((1 - x(1+ |1-x | )) /(|1-x|)) cos(1/(1-x))` for `x!=1`

Explanation

To solve the problem, we need to analyze the function \( f(x) = \frac{(1 - x)(1 + |1 - x|)}{|1 - x|} \cos\left(\frac{1}{1 - x}\right) \) for \( x \neq 1 \) and find the limits as \( x \) approaches 1 from the right (1+) and from the left (1-).

### Step 1: Calculate the limit as \( x \to 1^+ \)

Let \( x = 1 + h \) where \( h \to 0^+ \). Then we have:

\[ f(1 + h) = \frac{(1 - (1 + h))(1 + |1 - (1 + h)|)}{|1 - (1 + h)|} \cos\left(\frac{1}{1 - (1 + h)}\right) \]

This simplifies to:

\[ f(1 + h) = \frac{(-h)(1 + | - h|)}{| - h|} \cos\left(\frac{1}{-h}\right) \]

Since \( h \) is positive, \( | - h| = h \). Thus:

\[ f(1 + h) = \frac{(-h)(1 + h)}{h} \cos\left(-\frac{1}{h}\right) \]

This simplifies further to:

\[ f(1 + h) = -(1 + h) \cos\left(\frac{1}{h}\right) \]

Now, we need to find:

\[ \lim_{h \to 0^+} -(1 + h) \cos\left(\frac{1}{h}\right) \]

As \( h \to 0^+ \), \( -(1 + h) \) approaches -1, and \( \cos\left(\frac{1}{h}\right) \) oscillates between -1 and 1. Therefore, the limit does not exist.

### Step 2: Calculate the limit as \( x \to 1^- \)

Let \( x = 1 - h \) where \( h \to 0^+ \). Then we have:

\[ f(1 - h) = \frac{(1 - (1 - h))(1 + |1 - (1 - h)|)}{|1 - (1 - h)|} \cos\left(\frac{1}{1 - (1 - h)}\right) \]

This simplifies to:

\[ f(1 - h) = \frac{h(1 + |h|)}{|h|} \cos\left(\frac{1}{h}\right) \]

Since \( h \) is positive, \( |h| = h \). Thus:

\[ f(1 - h) = \frac{h(1 + h)}{h} \cos\left(\frac{1}{h}\right) = (1 + h) \cos\left(\frac{1}{h}\right) \]

Now, we need to find:

\[ \lim_{h \to 0^+} (1 + h) \cos\left(\frac{1}{h}\right) \]

As \( h \to 0^+ \), \( (1 + h) \) approaches 1, and \( \cos\left(\frac{1}{h}\right) \) oscillates between -1 and 1. Therefore, the limit approaches 0.

### Final Results

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

### Summary

Thus, the final answer is:
- \( \lim_{x \to 1^+} f(x) \) does not exist.
- \( \lim_{x \to 1^-} f(x) = 0 \).