The value of `lim_(xrarr0) (x(5^x-1))/(1-cos x)`, is

Options

A. `5log2`
B. `2log 5`
C. `(1)/(2)log 5`
D. `(1)/(5) log 2`

Answer

To solve the limit \( \lim_{x \to 0} \frac{x(5^x - 1)}{1 - \cos x} \), we will follow these steps:

### Step 1: Rewrite the limit
We start with the limit:
\[ L = \lim_{x \to 0} \frac{x(5^x - 1)}{1 - \cos x} \]

### Step 2: Apply L'Hôpital's Rule
As \( x \to 0 \), both the numerator and denominator approach 0. Therefore, we can apply L'Hôpital's Rule, which states that if the limit results in an indeterminate form \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator.

### Step 3: Differentiate the numerator and denominator
The derivative of the numerator \( x(5^x - 1) \) using the product rule is:
\[ \frac{d}{dx}[x(5^x - 1)] = (5^x - 1) + x \cdot 5^x \ln(5) \]
The derivative of the denominator \( 1 - \cos x \) is:
\[ \frac{d}{dx}[1 - \cos x] = \sin x \]

### Step 4: Rewrite the limit with derivatives
Now we can rewrite the limit using the derivatives:
\[ L = \lim_{x \to 0} \frac{(5^x - 1) + x \cdot 5^x \ln(5)}{\sin x} \]

### Step 5: Evaluate the limit
As \( x \to 0 \):
- \( 5^x - 1 \) approaches \( 0 \)
- \( x \cdot 5^x \ln(5) \) approaches \( 0 \)
- \( \sin x \) approaches \( 0 \)

Thus, we can apply L'Hôpital's Rule again. We differentiate the numerator and denominator again.

The derivative of the numerator:
\[ \frac{d}{dx}[(5^x - 1) + x \cdot 5^x \ln(5)] = 5^x \ln(5) + (5^x \ln(5) + x \cdot 5^x \ln^2(5)) \]
The derivative of the denominator \( \sin x \) is \( \cos x \).

### Step 6: Rewrite the limit again
Now we have:
\[ L = \lim_{x \to 0} \frac{5^x \ln(5) + 5^x \ln(5) + x \cdot 5^x \ln^2(5)}{\cos x} \]

### Step 7: Evaluate the limit as \( x \to 0 \)
As \( x \to 0 \):
- \( 5^x \) approaches \( 1 \)
- \( \cos x \) approaches \( 1 \)

Thus,
\[ L = \frac{2 \ln(5)}{1} = 2 \ln(5) \]

### Final Answer
The value of the limit is:
\[ \lim_{x \to 0} \frac{x(5^x - 1)}{1 - \cos x} = 2 \ln(5) \]