`lim_(x to 1/2)((8x-3)/(2x-1)-(4x^(2)+1)/(4x^(2)-1))`.

Answer

To solve the limit \( \lim_{x \to \frac{1}{2}} \left( \frac{8x - 3}{2x - 1} - \frac{4x^2 + 1}{4x^2 - 1} \right) \), we will follow these steps:

### Step 1: Substitute \( x = \frac{1}{2} \)
First, we substitute \( x = \frac{1}{2} \) into both fractions to check if we get an indeterminate form.

1. For the first term:
\[ 8 \left( \frac{1}{2} \right) - 3 = 4 - 3 = 1 \]
\[ 2 \left( \frac{1}{2} \right) - 1 = 1 - 1 = 0 \]
Thus, the first term becomes \( \frac{1}{0} \), which is undefined.

2. For the second term:
\[ 4 \left( \frac{1}{2} \right)^2 + 1 = 4 \cdot \frac{1}{4} + 1 = 1 + 1 = 2 \]
\[ 4 \left( \frac{1}{2} \right)^2 - 1 = 4 \cdot \frac{1}{4} - 1 = 1 - 1 = 0 \]
Thus, the second term also becomes \( \frac{2}{0} \), which is also undefined.

Since we have the form \( \infty - \infty \), we need to simplify the expression.

### Step 2: Find a common denominator
To combine the two fractions, we will find a common denominator:
\[ \text{Common Denominator} = (2x - 1)(4x^2 - 1) \]

### Step 3: Rewrite the expression
Rewriting the limit expression with the common denominator:
\[ \lim_{x \to \frac{1}{2}} \left( \frac{(8x - 3)(4x^2 - 1) - (4x^2 + 1)(2x - 1)}{(2x - 1)(4x^2 - 1)} \right) \]

### Step 4: Simplify the numerator
Now we will expand the numerator:
1. Expand \( (8x - 3)(4x^2 - 1) \):
\[ = 32x^3 - 8x - 12x^2 + 3 = 32x^3 - 12x^2 - 8x + 3 \]

2. Expand \( (4x^2 + 1)(2x - 1) \):
\[ = 8x^3 - 4x^2 + 2x - 1 \]

Now, combine these:
\[ 32x^3 - 12x^2 - 8x + 3 - (8x^3 - 4x^2 + 2x - 1) \]
\[ = 32x^3 - 8x^3 - 12x^2 + 4x^2 - 8x - 2x + 3 + 1 \]
\[ = 24x^3 - 8x^2 - 10x + 4 \]

### Step 5: Factor the numerator
Now we need to factor \( 24x^3 - 8x^2 - 10x + 4 \). We can factor out 2:
\[ = 2(12x^3 - 4x^2 - 5x + 2) \]

We can use synthetic division or factorization techniques to factor \( 12x^3 - 4x^2 - 5x + 2 \).

### Step 6: Evaluate the limit
Once we have factored the numerator, we can cancel out the common terms with the denominator \( (2x - 1)(4x^2 - 1) \).

After canceling, we can substitute \( x = \frac{1}{2} \) again to find the limit.

### Final Calculation
After simplification, we find:
\[ \lim_{x \to \frac{1}{2}} \frac{2(3x + 2)(2x - 1)}{(2x - 1)(2x + 1)} = \lim_{x \to \frac{1}{2}} \frac{2(3x + 2)}{2x + 1} \]
Substituting \( x = \frac{1}{2} \):
\[ = \frac{2(3 \cdot \frac{1}{2} + 2)}{2 \cdot \frac{1}{2} + 1} = \frac{2(\frac{3}{2} + 2)}{1 + 1} = \frac{2(\frac{7}{2})}{2} = 7 \]

Thus, the final answer is:
\[ \frac{7}{2} \]