`Lt_(x rarr oo) 2^(x) sin ((p)/(2^(x))) ,p in R ` equal to

Options

A. p log 2
B. p
C. `-p log 2`
D. `-p`

Answer

To solve the limit \( \lim_{x \to \infty} 2^x \sin\left(\frac{p}{2^x}\right) \), we can follow these steps:

### Step 1: Analyze the expression
As \( x \) approaches infinity, \( 2^x \) approaches infinity and \( \frac{p}{2^x} \) approaches 0. Therefore, we can rewrite the limit as:
\[ \lim_{x \to \infty} 2^x \sin\left(\frac{p}{2^x}\right) \]
This results in an indeterminate form of \( \infty \cdot 0 \).

### Step 2: Rewrite the limit
To resolve the indeterminate form, we can rewrite the expression:
\[ \lim_{x \to \infty} \frac{\sin\left(\frac{p}{2^x}\right)}{\frac{1}{2^x}} \]
Now, as \( x \to \infty \), both the numerator \( \sin\left(\frac{p}{2^x}\right) \) and the denominator \( \frac{1}{2^x} \) approach 0, creating a \( \frac{0}{0} \) indeterminate form.

### Step 3: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
if the limit exists.

#### Differentiate the numerator:
Using the chain rule:
\[ \frac{d}{dx} \left(\sin\left(\frac{p}{2^x}\right)\right) = \cos\left(\frac{p}{2^x}\right) \cdot \left(-\frac{p \ln(2)}{2^x}\right) \]

#### Differentiate the denominator:
\[ \frac{d}{dx} \left(\frac{1}{2^x}\right) = -\frac{\ln(2)}{2^x} \]

### Step 4: Substitute back into the limit
Now substituting back into the limit:
\[ \lim_{x \to \infty} \frac{\cos\left(\frac{p}{2^x}\right) \cdot \left(-\frac{p \ln(2)}{2^x}\right)}{-\frac{\ln(2)}{2^x}} = \lim_{x \to \infty} \frac{p \cos\left(\frac{p}{2^x}\right)}{1} \]

### Step 5: Evaluate the limit
As \( x \to \infty \), \( \frac{p}{2^x} \) approaches 0, thus \( \cos\left(\frac{p}{2^x}\right) \) approaches \( \cos(0) = 1 \). Therefore:
\[ \lim_{x \to \infty} p \cos\left(\frac{p}{2^x}\right) = p \cdot 1 = p \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to \infty} 2^x \sin\left(\frac{p}{2^x}\right) = p \]