Evaluate the following limits :
`Lim_(x to1/2)((8x-3)/(2x-1)-(4x^(2)+1)/(4x^(2)-1))`

Answer

To evaluate 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: Rewrite the limit expression
We start with the limit expression:

\[ \lim_{x \to \frac{1}{2}} \left( \frac{8x - 3}{2x - 1} - \frac{4x^2 + 1}{4x^2 - 1} \right). \]

### Step 2: Simplify the second term
The denominator \(4x^2 - 1\) can be factored using the difference of squares:

\[ 4x^2 - 1 = (2x - 1)(2x + 1). \]

Thus, we rewrite the second term:

\[ \frac{4x^2 + 1}{4x^2 - 1} = \frac{4x^2 + 1}{(2x - 1)(2x + 1)}. \]

### Step 3: Find a common denominator
Now we find a common denominator for the two fractions:

\[ \text{Common Denominator} = (2x - 1)(2x + 1). \]

Rewriting the limit expression with the common denominator gives us:

\[ \lim_{x \to \frac{1}{2}} \frac{(8x - 3)(2x + 1) - (4x^2 + 1)(2x - 1)}{(2x - 1)(2x + 1)}. \]

### Step 4: Expand the numerator
Now we expand the numerator:

1. Expanding \((8x - 3)(2x + 1)\):

\[ = 16x^2 + 8x - 6x - 3 = 16x^2 + 2x - 3. \]

2. Expanding \((4x^2 + 1)(2x - 1)\):

\[ = 8x^3 - 4x^2 + 2x - 1. \]

Now, substituting these expansions back into the limit gives:

\[ \lim_{x \to \frac{1}{2}} \frac{(16x^2 + 2x - 3) - (8x^3 - 4x^2 + 2x - 1)}{(2x - 1)(2x + 1)}. \]

### Step 5: Combine like terms
Now combine the terms in the numerator:

\[ = \lim_{x \to \frac{1}{2}} \frac{16x^2 + 2x - 3 - 8x^3 + 4x^2 - 2x + 1}{(2x - 1)(2x + 1)}. \]

This simplifies to:

\[ = \lim_{x \to \frac{1}{2}} \frac{-8x^3 + 20x^2 - 2}{(2x - 1)(2x + 1)}. \]

### Step 6: Factor the numerator
Factoring the numerator if possible or simplifying further. We can factor out \(-2\):

\[ = \lim_{x \to \frac{1}{2}} \frac{-2(4x^3 - 10x^2 + 1)}{(2x - 1)(2x + 1)}. \]

### Step 7: Evaluate the limit
Now substitute \(x = \frac{1}{2}\):

1. The denominator becomes:

\[ (2(\frac{1}{2}) - 1)(2(\frac{1}{2}) + 1) = (0)(2) = 0. \]

2. The numerator becomes:

\[ -2(4(\frac{1}{2})^3 - 10(\frac{1}{2})^2 + 1) = -2(4(\frac{1}{8}) - 10(\frac{1}{4}) + 1) = -2(\frac{1}{2} - \frac{5}{2} + 1) = -2(-2) = 4. \]

Thus, we have a \(0/0\) form, indicating we may need to apply L'Hôpital's Rule or further simplify.

### Step 8: Cancel common factors
Notice that \(2x - 1\) cancels out from the numerator and denominator, leading us to:

\[ \lim_{x \to \frac{1}{2}} \frac{4x + 2}{2x + 1}. \]

### Step 9: Final evaluation
Now substituting \(x = \frac{1}{2}\):

\[ \frac{4(\frac{1}{2}) + 2}{2(\frac{1}{2}) + 1} = \frac{2 + 2}{1 + 1} = \frac{4}{2} = 2. \]

Thus, the final answer is:

\[ \boxed{2}. \]