Evaluate the following limits:
`lim_(xrarr1/2)((4x^(2)-1)/(2x-1))`

Options

A. `-1`
B. `1`
C. `2`
D. `0`

Answer

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 we can directly find the limit.

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

This gives us an indeterminate form \( \frac{0}{0} \), so we need to simplify the expression.

### Step 2: Factor the numerator
We can factor the numerator \( 4x^2 - 1 \) using the difference of squares formula:

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

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

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

### Step 4: Cancel the common terms
Since \( 2x - 1 \) is in both the numerator and the denominator, we can cancel it (as long as \( x \neq \frac{1}{2} \)):

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

### Step 5: 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:

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

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