Let `f(x)=(1)/(1-|1-x|)`. Then, what is `lim_(x to 0) f(x)` equal to

Options

A. 0
B. `oo`
C. 1
D. -1

Explanation

To find the limit of the function \( f(x) = \frac{1}{1 - |1 - x|} \) as \( x \) approaches 0, we will evaluate both the left-hand limit (LHL) and the right-hand limit (RHL) at \( x = 0 \).

### Step 1: Evaluate the Left-Hand Limit (LHL)
We start by calculating the left-hand limit as \( x \) approaches 0 from the left (denoted as \( x \to 0^- \)).

1. **Substituting \( x = 0 - h \)** (where \( h \) is a small positive number):
\[ f(0 - h) = \frac{1}{1 - |1 - (0 - h)|} = \frac{1}{1 - |1 - 0 + h|} = \frac{1}{1 - |1 + h|} \]
Since \( h \) is positive, \( |1 + h| = 1 + h \):
\[ f(0 - h) = \frac{1}{1 - (1 + h)} = \frac{1}{1 - 1 - h} = \frac{1}{-h} \]

2. **Taking the limit as \( h \to 0^+ \)**:
\[ \lim_{h \to 0^+} f(0 - h) = \lim_{h \to 0^+} \frac{1}{-h} = -\infty \]

### Step 2: Evaluate the Right-Hand Limit (RHL)
Next, we calculate the right-hand limit as \( x \) approaches 0 from the right (denoted as \( x \to 0^+ \)).

1. **Substituting \( x = 0 + h \)** (where \( h \) is a small positive number):
\[ f(0 + h) = \frac{1}{1 - |1 - (0 + h)|} = \frac{1}{1 - |1 - 0 - h|} = \frac{1}{1 - |1 - h|} \]
Since \( h \) is positive and small, \( |1 - h| = 1 - h \):
\[ f(0 + h) = \frac{1}{1 - (1 - h)} = \frac{1}{1 - 1 + h} = \frac{1}{h} \]

2. **Taking the limit as \( h \to 0^+ \)**:
\[ \lim_{h \to 0^+} f(0 + h) = \lim_{h \to 0^+} \frac{1}{h} = +\infty \]

### Step 3: Conclusion
Since the left-hand limit and the right-hand limit are not equal:
\[ \lim_{x \to 0^-} f(x) = -\infty \quad \text{and} \quad \lim_{x \to 0^+} f(x) = +\infty \]
Thus, the overall limit does not exist:
\[ \lim_{x \to 0} f(x) \text{ does not exist.} \]