Find the value of the limit given below
`lim_(n to 1/2)(4n^(2)-1)/(2n-1)`

Answer

To find the limit

\[ \lim_{n \to \frac{1}{2}} \frac{4n^2 - 1}{2n - 1}, \]

we will follow these steps:

### Step 1: Substitute the limit value
First, we will substitute \( n = \frac{1}{2} \) into the expression to check if we get an indeterminate form.

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

Since both the numerator and denominator approach 0, we have the indeterminate form \( \frac{0}{0} \).

### Step 2: Factor the numerator
Next, we will factor the numerator \( 4n^2 - 1 \). We can recognize this as a difference of squares:

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

### Step 3: Rewrite the limit
Now we can rewrite the limit using the factored form of the numerator:

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

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

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

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

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

### Conclusion
Thus, the value of the limit is

\[ \boxed{2}. \]