Evaluate `lim_(xto 1//2) ((4x^(2)-1))/((2x-1))`.

Options

A. 2
B. 3
C. 0
D. 1

Explanation

To evaluate the limit \( \lim_{x \to \frac{1}{2}} \frac{4x^2 - 1}{2x - 1} \), we will follow these steps:

### Step 1: Substitute \( x = \frac{1}{2} \) into the expression
First, we can substitute \( x = \frac{1}{2} \) directly into the expression to see if we get a determinate form.

\[ 4\left(\frac{1}{2}\right)^2 - 1 = 4 \cdot \frac{1}{4} - 1 = 1 - 1 = 0 \]
\[ 2\left(\frac{1}{2}\right) - 1 = 1 - 1 = 0 \]

Since both the numerator and denominator evaluate to 0, we have an indeterminate form \( \frac{0}{0} \). We need to simplify the expression.

### Step 2: Factor the numerator
The numerator \( 4x^2 - 1 \) can be factored using the difference of squares:

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

Now, we can rewrite the limit:

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

### Step 3: Cancel the common factors
Since \( 2x - 1 \) is a common factor in both the numerator and the denominator, we can cancel it out (as long as \( x \neq \frac{1}{2} \)):

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

### Step 4: Substitute \( x = \frac{1}{2} \) again
Now we can substitute \( x = \frac{1}{2} \) into the simplified expression:

\[ 2\left(\frac{1}{2}\right) + 1 = 1 + 1 = 2 \]

### Final Answer
Thus, the limit is:

\[ \lim_{x \to \frac{1}{2}} \frac{4x^2 - 1}{2x - 1} = 2 \]