`lim_(xrarr 1) (x^(2^(32))-2^32x+4^16-1)/((x-1)^2)` is equal to

Options

A. `2^63-2^31`
B. `2^64-2^31`
C. `2^62-2^31`
D. `2^65- 2^33`

Explanation

To solve the limit

\[ L = \lim_{x \to 1} \frac{x^{2^{32}} - 2^{32}x + 4^{16} - 1}{(x-1)^2} \]

we will follow these steps:

### Step 1: Simplify the expression
We can rewrite \(4^{16}\) as \( (2^2)^{16} = 2^{32} \). Thus, the limit becomes:

\[ L = \lim_{x \to 1} \frac{x^{2^{32}} - 2^{32}x + 2^{32} - 1}{(x-1)^2} \]

### Step 2: Substitute \(n = 2^{32}\)
Let \(n = 2^{32}\). The limit now looks like:

\[ L = \lim_{x \to 1} \frac{x^n - nx + n - 1}{(x-1)^2} \]

### Step 3: Factor the numerator
The numerator can be factored using the identity \(x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \ldots + 1)\):

\[ x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \ldots + 1) \]

Thus, we can rewrite \(x^n - nx + n - 1\) as:

\[ L = \lim_{x \to 1} \frac{(x-1)(x^{n-1} + x^{n-2} + \ldots + 1 - n)}{(x-1)^2} \]

### Step 4: Cancel out \((x-1)\)
We can cancel one \((x-1)\) from the numerator and denominator:

\[ L = \lim_{x \to 1} \frac{x^{n-1} + x^{n-2} + \ldots + 1 - n}{x-1} \]

### Step 5: Evaluate the limit
Now we can evaluate the limit by substituting \(x = 1\):

\[ L = \lim_{x \to 1} \left( (n-1) + (n-2) + \ldots + 1 \right) \]

This is the sum of the first \(n-1\) natural numbers, which can be calculated using the formula:

\[ \text{Sum} = \frac{(n-1)n}{2} \]

### Step 6: Substitute back \(n = 2^{32}\)
Now substituting back \(n = 2^{32}\):

\[ L = \frac{(2^{32}-1) \cdot 2^{32}}{2} \]

### Step 7: Simplify the expression
This simplifies to:

\[ L = 2^{32} \cdot (2^{32} - 1) / 2 = \frac{2^{64} - 2^{32}}{2} = 2^{63} - 2^{31} \]

Thus, the final answer is:

\[ \boxed{2^{63} - 2^{31}} \]