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 can follow these steps:

### Step 1: Substitute \( x = \frac{1}{2} \)
First, we substitute \( x = \frac{1}{2} \) into the expression to check if it results in an indeterminate 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 the indeterminate form \( \frac{0}{0} \).

**Hint:** When you encounter \( \frac{0}{0} \), it indicates that further simplification or factorization is needed.

### Step 2: Factor the numerator
Next, we factor the numerator \( 4x^2 - 1 \). This expression can be recognized as a difference of squares:

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

**Hint:** Remember that the difference of squares can be factored as \( a^2 - b^2 = (a - b)(a + b) \).

### Step 3: Rewrite the limit
Now we can rewrite the limit using the factorization:

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

**Hint:** When you factor out common terms, it can help eliminate the indeterminate form.

### Step 4: Cancel the common factor
We can cancel the common factor \( 2x - 1 \) from the numerator and denominator:

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

**Hint:** Always check if you can simplify the expression before substituting values.

### Step 5: Substitute again
Now we can substitute \( x = \frac{1}{2} \) into the simplified expression:

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

**Hint:** After simplification, substituting the limit point should yield a definite value.

### Final Answer
Thus, the limit evaluates to:

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