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To solve this limit problem, we need to find:

\[
\lim_{{x \to 0}} \frac{\cos(3x) - 1}{5x + \tan(x)}
\]

### Step-by-Step Solution

1. **Evaluate the form**:
   As \( x \to 0 \), we can see that \( \cos(3x) \to 1 \), \( \tan(x) \to 0 \), and \(5x \to 0\). So, the expression takes the indeterminate form \( \frac{0}{0} \).

2. **Apply L'Hôpital's Rule**:
   Since the limit is an indeterminate form of type \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that:
   \[
   \lim_{{x \to 0}} \frac{f(x)}{g(x)} = \lim_{{x \to 0}} \frac{f'(x)}{g'(x)}
   \]
   if the limit exists. Here:
   - \( f(x) = \cos(3x) - 1 \)
   - \( g(x) = 5x + \tan(x) \)

3. **Differentiate the Numerator** \( f(x) \):
   \[
   f'(x) = \frac{d}{dx}(\cos(3x) - 1) = -3 \sin(3x)
   \]

4. **Differentiate the Denominator** \( g(x) \):
   \[
   g'(x) = \frac{d}{dx}(5x + \tan(x)) = 5 + \sec^2(x)
   \]

5. **Rewrite the Limit**:
   Substitute \( f'(x) \) and \( g'(x) \) back into the limit:
   \[
   \lim_{{x \to 0}} \frac{\cos(3x) - 1}{5x + \tan(x)} = \lim_{{x \to 0}} \frac{-3 \sin(3x)}{5 + \sec^2(x)}
   \]

6. **Evaluate the New Limit**:
   As \( x \to 0 \):
   - \( \sin(3x) \approx 3x \), so \( -3 \sin(3x) \approx -9x \).
   - \( \sec^2(x) \to 1 \), so \( 5 + \sec^2(x) \to 6 \).

   This gives:
   \[
   \lim_{{x \to 0}} \frac{-9x}{6} = \frac{-9}{6} x = 0
   \]


Find the limit as x approaches 0 of (cos(3x) - 1) / (5x + tan(x))

Solve the limit of (cos(3x) - 1) / (5x + tan(x)) as x approaches 0 using L'Hôpital's Rule to address the indeterminate form 0/0. This problem applies calculus concepts, including trigonometric limits, making it suitable for Grades 11-12.

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