If f(x) `{:(=(x^(2)-4x+3)/(x^(2)-1)", ..."x!=1 ),( =2", ..."x=1):}` then :

Options

A. ` lim_(x to 1^(+) ) f(x) = 2 `
B. ` lim _(x to 1^(-)) f(x) = 3 `
C. f(x) is dicontinuous at x = 1
D. none of these

Answer

To determine the continuity of the function \( f(x) \) at \( x = 1 \), we need to follow these steps:

### Step 1: Identify the function and the point of continuity
The function is defined as follows:
\[ f(x) = \begin{cases} \frac{x^2 - 4x + 3}{x^2 - 1} & \text{if } x \neq 1 \\ 2 & \text{if } x = 1 \end{cases} \]
We need to check the continuity at \( x = 1 \).

### Step 2: Find \( f(1) \)
From the definition of the function:
\[ f(1) = 2 \]

### Step 3: Calculate the limit as \( x \) approaches 1
We need to find:
\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x^2 - 4x + 3}{x^2 - 1} \]

### Step 4: Factor the numerator and denominator
First, we factor the numerator \( x^2 - 4x + 3 \) and the denominator \( x^2 - 1 \):
- The numerator can be factored as:
\[ x^2 - 4x + 3 = (x - 3)(x - 1) \]
- The denominator can be factored as:
\[ x^2 - 1 = (x - 1)(x + 1) \]

### Step 5: Rewrite the limit
Now we can rewrite the limit:
\[ \lim_{x \to 1} \frac{(x - 3)(x - 1)}{(x - 1)(x + 1)} \]
Notice that \( (x - 1) \) cancels out:
\[ \lim_{x \to 1} \frac{x - 3}{x + 1} \]

### Step 6: Evaluate the limit
Now we can substitute \( x = 1 \) into the simplified expression:
\[ \frac{1 - 3}{1 + 1} = \frac{-2}{2} = -1 \]

### Step 7: Compare the limit to \( f(1) \)
We found:
\[ \lim_{x \to 1} f(x) = -1 \quad \text{and} \quad f(1) = 2 \]

### Conclusion
Since \( \lim_{x \to 1} f(x) \neq f(1) \), the function \( f(x) \) is not continuous at \( x = 1 \).

---