`lim_(xrarr oo) (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 \( \lim_{x \to 1} \frac{x^{2^{32}} - 2^{32}x + 4^{16} - 1}{(x-1)^2} \), we will follow these steps:

### Step 1: Rewrite the expression
First, we notice that \( 4^{16} = (2^2)^{16} = 2^{32} \). Thus, we can rewrite the limit as:
\[ \lim_{x \to 1} \frac{x^{2^{32}} - 2^{32}x + 2^{32} - 1}{(x-1)^2} \]

### Step 2: Substitute \( n = 2^{32} \)
Let \( n = 2^{32} \). The limit now becomes:
\[ \lim_{x \to 1} \frac{x^n - nx + n - 1}{(x-1)^2} \]

### Step 3: Factor the numerator
We can factor the numerator:
\[ x^n - nx + n - 1 = (x^n - 1) - n(x - 1) \]
Thus, we rewrite the limit:
\[ \lim_{x \to 1} \frac{(x^n - 1) - n(x - 1)}{(x-1)^2} \]

### Step 4: Apply L'Hôpital's Rule
Since both the numerator and denominator approach 0 as \( x \to 1 \), we can apply L'Hôpital's Rule. We differentiate the numerator and denominator:
- The derivative of the numerator \( (x^n - 1 - n(x - 1)) \) is \( nx^{n-1} - n \).
- The derivative of the denominator \( (x-1)^2 \) is \( 2(x-1) \).

Now we have:
\[ \lim_{x \to 1} \frac{nx^{n-1} - n}{2(x-1)} \]

### Step 5: Apply L'Hôpital's Rule again
As \( x \to 1 \), the numerator still approaches 0. We apply L'Hôpital's Rule again:
- The derivative of the numerator \( nx^{n-1} - n \) is \( n(n-1)x^{n-2} \).
- The derivative of the denominator \( 2(x-1) \) is \( 2 \).

Now we have:
\[ \lim_{x \to 1} \frac{n(n-1)x^{n-2}}{2} \]

### Step 6: Evaluate the limit
Substituting \( x = 1 \):
\[ \frac{n(n-1)}{2} = \frac{2^{32}(2^{32} - 1)}{2} = 2^{31}(2^{32} - 1) \]

### Final Result
Thus, the limit evaluates to:
\[ 2^{31}(2^{32} - 1) = 2^{63} - 2^{31} \]

### Answer
The limit is:
\[ \boxed{2^{63} - 2^{31}} \]