`lim_(x to 1//2) (8x^(3) - 1)/(16 x^(4) - 1)` is equal to

Options

A. `(1)/(2)`
B. `(3)/(2)`
C. `(3)/(4)`
D. `(1)/(4)`

Answer

To solve the limit \( \lim_{x \to \frac{1}{2}} \frac{8x^3 - 1}{16x^4 - 1} \), we will follow these steps:

### Step 1: Factor the numerator and denominator
We start by factoring the expressions in the numerator and denominator.

The numerator \( 8x^3 - 1 \) can be factored using the difference of cubes:
\[ 8x^3 - 1 = (2x - 1)(4x^2 + 2x + 1) \]

The denominator \( 16x^4 - 1 \) can be factored using the difference of squares:
\[ 16x^4 - 1 = (4x^2 - 1)(4x^2 + 1) = (2x - 1)(2x + 1)(4x^2 + 1) \]

### Step 2: Rewrite the limit
Now we can rewrite the limit using the factored forms:
\[ \lim_{x \to \frac{1}{2}} \frac{(2x - 1)(4x^2 + 2x + 1)}{(2x - 1)(2x + 1)(4x^2 + 1)} \]

### Step 3: Cancel common factors
We can cancel the common factor \( (2x - 1) \) from the numerator and the denominator:
\[ \lim_{x \to \frac{1}{2}} \frac{4x^2 + 2x + 1}{(2x + 1)(4x^2 + 1)} \]

### Step 4: Substitute \( x = \frac{1}{2} \)
Now we substitute \( x = \frac{1}{2} \) into the simplified expression:
\[ = \frac{4\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) + 1}{(2\left(\frac{1}{2}\right) + 1)(4\left(\frac{1}{2}\right)^2 + 1)} \]

Calculating the numerator:
\[ = 4 \cdot \frac{1}{4} + 1 + 1 = 1 + 1 + 1 = 3 \]

Calculating the denominator:
\[ = (1 + 1)(1 + 1) = 2 \cdot 2 = 4 \]

### Step 5: Final result
Thus, we have:
\[ \lim_{x \to \frac{1}{2}} \frac{8x^3 - 1}{16x^4 - 1} = \frac{3}{4} \]

### Final Answer
\[ \frac{3}{4} \]