If `f(x)=(x.2^(x)-x)/(1-cosx)` and `g(x)=2^(x).sin((log2)/(2^(x)))` then

Options

A. `lim_(xto0)f(x)=log2`
B. `lim_(xtooo)g(x)=log4`
C. `lim_(xto0)f(x)=log4`
D. `lim_(xtooo)g(x)=log2`

Answer

To solve the problem, we need to find the limits of the two functions \( f(x) \) and \( g(x) \) as \( x \) approaches 0 and infinity, respectively.

### Step 1: Evaluate \( \lim_{x \to 0} f(x) \)

Given:
\[ f(x) = \frac{x \cdot 2^x - x}{1 - \cos x} \]
First, substitute \( x = 0 \):
\[ f(0) = \frac{0 \cdot 2^0 - 0}{1 - \cos(0)} = \frac{0}{0} \text{ (indeterminate form)} \]

### Step 2: Simplify \( f(x) \)

We can factor out \( x \) from the numerator:
\[ f(x) = \frac{x(2^x - 1)}{1 - \cos x} \]

### Step 3: Rewrite the denominator

Using the identity \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \):
\[ f(x) = \frac{x(2^x - 1)}{2 \sin^2\left(\frac{x}{2}\right)} \]

### Step 4: Apply limits

Now, we can separate the limit:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x(2^x - 1)}{2 \sin^2\left(\frac{x}{2}\right)} \]

### Step 5: Use known limits

Using the known limits:
1. \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)
2. \( \lim_{x \to 0} \frac{2^x - 1}{x} = \ln(2) \)

We can rewrite the limit:
\[ \lim_{x \to 0} f(x) = \frac{1}{2} \cdot \lim_{x \to 0} \frac{x(2^x - 1)}{\sin^2\left(\frac{x}{2}\right)} \]

### Step 6: Substitute the limits

Substituting the limits:
\[ \lim_{x \to 0} f(x) = \frac{1}{2} \cdot \ln(2) \cdot \lim_{x \to 0} \frac{x}{\left(\frac{x}{2}\right)^2} \]
This simplifies to:
\[ \lim_{x \to 0} f(x) = \frac{1}{2} \cdot \ln(2) \cdot \frac{4}{1} = 2 \ln(2) \]

### Step 7: Final result for \( f(x) \)

Thus:
\[ \lim_{x \to 0} f(x) = \ln(4) \]

### Step 8: Evaluate \( \lim_{x \to \infty} g(x) \)

Given:
\[ g(x) = 2^x \cdot \sin\left(\frac{\log 2}{2^x}\right) \]

### Step 9: Substitute \( x \to \infty \)

As \( x \to \infty \), \( \frac{\log 2}{2^x} \to 0 \):
\[ g(x) = 2^x \cdot \sin\left(\frac{\log 2}{2^x}\right) \to \infty \cdot 0 \text{ (indeterminate form)} \]

### Step 10: Rewrite using limits

We can rewrite:
\[ g(x) = \frac{\log 2}{2^x} \cdot \frac{2^x \cdot \sin\left(\frac{\log 2}{2^x}\right)}{\frac{\log 2}{2^x}} \]

### Step 11: Apply known limit

Using the limit:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
We have:
\[ \lim_{x \to \infty} g(x) = \log 2 \cdot 1 = \log 2 \]

### Final Results

1. \( \lim_{x \to 0} f(x) = \ln(4) \)
2. \( \lim_{x \to \infty} g(x) = \log(2) \)