Evaluate the following limits :
`Lim_(x to 1/2)(4x^(2)-1)/(2x-1)`

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 the limit value
First, we can substitute \(x = \frac{1}{2}\) directly into the expression.

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

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

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

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

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

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

### Step 4: Cancel common factors
Since \(2x - 1\) appears 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 5: Substitute the limit value again
Now we substitute \(x = \frac{1}{2}\) into the simplified expression:

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

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

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

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