Let `f(x) = ((x - 1))/((2x^(2) - 7x + 5))` then

Options

A. `underset(x to 1)(lim) f(x) = - (1)/(3)`
B. `underset(x to 0)(lim) f(x) = - (1)/(5)`
C. `underset(x to oo)(lim) f (x) = 0`
D. `underset(x + (5)/(2))(lim) f(x)` does not exist

Answer

To solve the problem step by step, we will analyze the function \( f(x) = \frac{x - 1}{2x^2 - 7x + 5} \) and find the limits as specified in the question.

### Step 1: Factor the denominator
We start with the function:
\[ f(x) = \frac{x - 1}{2x^2 - 7x + 5} \]
We need to factor the denominator \( 2x^2 - 7x + 5 \). To do this, we will find the roots of the quadratic equation.

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = -7, c = 5 \):
\[ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 5}}{2 \cdot 2} = \frac{7 \pm \sqrt{49 - 40}}{4} = \frac{7 \pm 3}{4} \]
The roots are:
\[ x_1 = \frac{10}{4} = \frac{5}{2}, \quad x_2 = \frac{4}{4} = 1 \]
Thus, we can factor the denominator as:
\[ 2x^2 - 7x + 5 = 2(x - 1)(x - \frac{5}{2}) \]

### Step 2: Simplify the function
Now substituting the factored form back into \( f(x) \):
\[ f(x) = \frac{x - 1}{2(x - 1)(x - \frac{5}{2})} \]
We can cancel \( (x - 1) \) from the numerator and denominator (for \( x \neq 1 \)):
\[ f(x) = \frac{1}{2(x - \frac{5}{2})}, \quad x \neq 1 \]

### Step 3: Find the limit as \( x \to 1 \)
Now we will find:
\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{1}{2(x - \frac{5}{2})} \]
Substituting \( x = 1 \):
\[ = \frac{1}{2(1 - \frac{5}{2})} = \frac{1}{2(-\frac{3}{2})} = -\frac{1}{3} \]

### Step 4: Find the limit as \( x \to 0 \)
Next, we find:
\[ \lim_{x \to 0} f(x) = \frac{0 - 1}{2(0^2) - 7(0) + 5} = \frac{-1}{5} \]

### Step 5: Find the limit as \( x \to \infty \)
Now we calculate:
\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{1}{2(x - \frac{5}{2})} \]
As \( x \to \infty \):
\[ = \frac{1}{2(\infty - \frac{5}{2})} = 0 \]

### Step 6: Find the limit as \( x \to \frac{5}{2} \)
Now we find the limit as \( x \to \frac{5}{2} \):
- Right-hand limit:
\[ \lim_{x \to \frac{5}{2}^+} f(x) = \frac{1}{2(0^+)} = +\infty \]
- Left-hand limit:
\[ \lim_{x \to \frac{5}{2}^-} f(x) = \frac{1}{2(0^-)} = -\infty \]
Since the left-hand limit and right-hand limit are not equal, the limit does not exist.

### Summary of Results
1. \( \lim_{x \to 1} f(x) = -\frac{1}{3} \)
2. \( \lim_{x \to 0} f(x) = -\frac{1}{5} \)
3. \( \lim_{x \to \infty} f(x) = 0 \)
4. \( \lim_{x \to \frac{5}{2}} f(x) \) does not exist.