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 will follow these steps:

### Step 1: Identify the Form of the Limit
First, we substitute \( x = 0 \) into the expression:
- \( \sin(5 \cdot 0) + \cos(5 \cdot 0) = \sin(0) + \cos(0) = 0 + 1 = 1 \)
- Thus, \( \log(1) = 0 \)
- For the denominator, \( \tan(3 \cdot 0) = \tan(0) = 0 \)

This gives us the indeterminate form \( \frac{0}{0} \).

**Hint:** When you encounter the \( \frac{0}{0} \) form, consider using L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
if the limit on the right exists.

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

- **Denominator:** \( g(x) = \tan(3x) \)
- The derivative is:
\[ g'(x) = 3\sec^2(3x) \]

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

### Step 5: Substitute \( x = 0 \)
Now we substitute \( x = 0 \) into the new limit:
- The numerator becomes:
\[ 5(\cos(0) - \sin(0)) = 5(1 - 0) = 5 \]
- The denominator becomes:
\[ (\sin(0) + \cos(0)) \cdot 3\sec^2(0) = (0 + 1) \cdot 3 \cdot 1 = 3 \]

### Step 6: Final Calculation
Now we can compute the limit:
\[ \lim_{x \to 0} \frac{5}{3} = \frac{5}{3} \]

### Conclusion
Thus, the value of the limit is:
\[ \boxed{\frac{5}{3}} \]