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: $\lim_{{x \to 0}} \frac{-9x}{6} = \frac{-9}{6} x = 0$