If `f:R to R` is defined by `f(x)={{:(,(x-2)/(x^(2)-3x+2),"if "x in R-(1,2)),(,2,"if "x=1),(,1,"if "x=2):}`
`"them " lim_(x to 2) (f(x)-f(2))/(x-2)=`

Options

A. 0
B. `-1`
C. 1
D. `-1//2`

Explanation

To solve the problem step-by-step, we need to find the limit:

\[ \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} \]

where the function \( f(x) \) is defined as:

\[ f(x) = \begin{cases} \frac{x - 2}{x^2 - 3x + 2} & \text{if } x \in \mathbb{R} \setminus (1, 2) \\ 2 & \text{if } x = 1 \\ 1 & \text{if } x = 2 \end{cases} \]

### Step 1: Identify \( f(2) \)

From the definition of the function, we see that:

\[ f(2) = 1 \]

### Step 2: Simplify \( f(x) \) for \( x \) near 2

For \( x \) approaching 2 (but not equal to 2), we use the first case of the function:

\[ f(x) = \frac{x - 2}{x^2 - 3x + 2} \]

### Step 3: Factor the denominator

The denominator can be factored:

\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]

Thus, we can rewrite \( f(x) \):

\[ f(x) = \frac{x - 2}{(x - 1)(x - 2)} \]

### Step 4: Cancel common terms

For \( x \neq 2 \), we can cancel \( x - 2 \):

\[ f(x) = \frac{1}{x - 1} \quad \text{for } x \in \mathbb{R} \setminus (1, 2) \]

### Step 5: Substitute into the limit expression

Now we substitute \( f(x) \) and \( f(2) \) into the limit:

\[ \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} = \lim_{x \to 2} \frac{\frac{1}{x - 1} - 1}{x - 2} \]

### Step 6: Find a common denominator

To simplify the expression in the limit, we find a common denominator:

\[ \frac{1}{x - 1} - 1 = \frac{1 - (x - 1)}{x - 1} = \frac{2 - x}{x - 1} \]

Thus, we have:

\[ \lim_{x \to 2} \frac{\frac{2 - x}{x - 1}}{x - 2} = \lim_{x \to 2} \frac{2 - x}{(x - 1)(x - 2)} \]

### Step 7: Substitute and simplify

Now we can substitute \( x = 2 \):

\[ = \lim_{x \to 2} \frac{-(x - 2)}{(x - 1)(x - 2)} = \lim_{x \to 2} \frac{-1}{x - 1} \]

### Step 8: Evaluate the limit

Now substituting \( x = 2 \):

\[ = \frac{-1}{2 - 1} = -1 \]

### Final Answer

Thus, the limit is:

\[ \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} = -1 \]