`"lt"_(x to 0)(8^(x) -4^(x) -2^(x) +1^(x))/x^(2)`=

Options

A. 2 log 2
B. `2(log 2)(2)`
C. log 2
D. none

Answer

To solve the limit \(\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1^x}{x^2}\), we will follow these steps:

### Step 1: Identify the form of the limit
First, we substitute \(x = 0\) into the expression:

\[ 8^0 - 4^0 - 2^0 + 1^0 = 1 - 1 - 1 + 1 = 0 \]

The denominator is:

\[ 0^2 = 0 \]

Thus, we have the indeterminate form \(\frac{0}{0}\).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator separately.

The numerator is \(8^x - 4^x - 2^x + 1^x\) and the denominator is \(x^2\).

### Step 3: Differentiate the numerator and denominator
Differentiating the numerator:

\[ \frac{d}{dx}(8^x) = 8^x \ln(8), \quad \frac{d}{dx}(4^x) = 4^x \ln(4), \quad \frac{d}{dx}(2^x) = 2^x \ln(2), \quad \frac{d}{dx}(1^x) = 0 \]

Thus, the derivative of the numerator is:

\[ 8^x \ln(8) - 4^x \ln(4) - 2^x \ln(2) \]

The derivative of the denominator is:

\[ \frac{d}{dx}(x^2) = 2x \]

### Step 4: Rewrite the limit
Now we rewrite the limit using the derivatives:

\[ \lim_{x \to 0} \frac{8^x \ln(8) - 4^x \ln(4) - 2^x \ln(2)}{2x} \]

### Step 5: Substitute \(x = 0\) again
Substituting \(x = 0\) into the new limit:

\[ 8^0 \ln(8) - 4^0 \ln(4) - 2^0 \ln(2) = \ln(8) - \ln(4) - \ln(2) = \ln\left(\frac{8}{4 \cdot 2}\right) = \ln(1) = 0 \]

The denominator becomes \(2 \cdot 0 = 0\) again, so we apply L'Hôpital's Rule a second time.

### Step 6: Differentiate again
We differentiate the numerator again:

\[ \frac{d}{dx}(8^x \ln(8)) = 8^x \ln(8)^2, \quad \frac{d}{dx}(4^x \ln(4)) = 4^x \ln(4)^2, \quad \frac{d}{dx}(2^x \ln(2)) = 2^x \ln(2)^2 \]

Thus, the new numerator is:

\[ 8^x \ln(8)^2 - 4^x \ln(4)^2 - 2^x \ln(2)^2 \]

The denominator's derivative is:

\[ \frac{d}{dx}(2x) = 2 \]

### Step 7: Rewrite the limit again
Now we rewrite the limit:

\[ \lim_{x \to 0} \frac{8^x \ln(8)^2 - 4^x \ln(4)^2 - 2^x \ln(2)^2}{2} \]

### Step 8: Substitute \(x = 0\) one last time
Substituting \(x = 0\):

\[ 8^0 \ln(8)^2 - 4^0 \ln(4)^2 - 2^0 \ln(2)^2 = \ln(8)^2 - \ln(4)^2 - \ln(2)^2 \]

Using the logarithmic identities:

\[ \ln(8) = 3\ln(2), \quad \ln(4) = 2\ln(2) \]

Thus:

\[ \ln(8)^2 = (3\ln(2))^2 = 9\ln(2)^2, \quad \ln(4)^2 = (2\ln(2))^2 = 4\ln(2)^2 \]

So:

\[ 9\ln(2)^2 - 4\ln(2)^2 - \ln(2)^2 = 4\ln(2)^2 \]

Finally, we have:

\[ \lim_{x \to 0} \frac{4\ln(2)^2}{2} = 2\ln(2)^2 \]

### Final Answer
Thus, the limit is:

\[ \boxed{2\ln(2)^2} \]