The value of `lim_(x to 0) (log(sin 5x + cos 5x))/(tan 3x)` is equal to

Options

A. `10/3`
B. `20/3`
C. `5/6`
D. `5/3`

Answer

To solve the limit \( \lim_{x \to 0} \frac{\log(\sin(5x) + \cos(5x))}{\tan(3x)} \), we can follow these steps:

### Step 1: Identify the form of the limit
As \( x \to 0 \):
- \( \sin(5x) \to 0 \)
- \( \cos(5x) \to 1 \)
Thus, \( \sin(5x) + \cos(5x) \to 0 + 1 = 1 \).
Therefore, \( \log(\sin(5x) + \cos(5x)) \to \log(1) = 0 \).

Also, \( \tan(3x) \to 0 \) as \( x \to 0 \).

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

### Step 2: 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 on the right exists.

### Step 3: Differentiate the numerator and denominator
1. **Differentiate the numerator**:
\[ f(x) = \log(\sin(5x) + \cos(5x)) \]
Using the chain rule:
\[ f'(x) = \frac{1}{\sin(5x) + \cos(5x)} \cdot (5\cos(5x) - 5\sin(5x)) \]
Simplifying:
\[ f'(x) = \frac{5(\cos(5x) - \sin(5x))}{\sin(5x) + \cos(5x)} \]

2. **Differentiate the denominator**:
\[ g(x) = \tan(3x) \]
The derivative is:
\[ g'(x) = 3\sec^2(3x) \]

### Step 4: Rewrite the limit using derivatives
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{5(\cos(5x) - \sin(5x))}{\sin(5x) + \cos(5x)} \cdot \frac{1}{3\sec^2(3x)} \]

### Step 5: Evaluate the limit
Substituting \( x = 0 \):
- \( \cos(5 \cdot 0) = 1 \)
- \( \sin(5 \cdot 0) = 0 \)
- Thus, \( \sin(5x) + \cos(5x) \to 1 \)

So we have:
\[ \lim_{x \to 0} \frac{5(1 - 0)}{1} \cdot \frac{1}{3 \cdot 1} = \frac{5}{3} \]

### Final Result
Therefore, the value of the limit is:
\[ \boxed{\frac{5}{3}} \]