To solve the limit

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

let's analyze the numerator and the denominator separately. Notice that as \(x \to 0\), both the numerator and the denominator approach 0, which is an indeterminate form of type \(\frac{0}{0}\). To resolve this, we can apply L'Hopital's rule.

### Step-by-step Solution

1. **Differentiate the numerator and denominator:**

   The numerator is \(f(x) = x + \cos(5x) - 1\), and the denominator is \(g(x) = -3x\).

   Find the derivative of the numerator:

   \[
   f'(x) = \frac{d}{dx} \left( x + \cos(5x) - 1 \right) = 1 - 5 \sin(5x).
   \]

   Find the derivative of the denominator:

   \[
   g'(x) = \frac{d}{dx}(-3x) = -3.
   \]

2. **Apply L'Hopital's rule:**

   Now, compute the limit:

   \[
   \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{1 - 5 \sin(5x)}{-3}.
   \]

3. **Evaluate the limit:**

   As \(x \to 0\), \(\sin(5x) \to 0\). Therefore,

   \[
   \lim_{x \to 0} \frac{1 - 5 \sin(5x)}{-3} = \frac{1 - 0}{-3} = -\frac{1}{3}.
   \]

### Final Answer

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

Would you like more details on this solution, or do you have any questions? 

Here are 5 related questions for further exploration:

1. How do you determine when to apply L'Hopital's rule?
2. What are other methods for solving limits involving trigonometric functions?
3. Can this limit be solved using Taylor series expansion?
4. How does the limit change if the function inside the cosine is different (e.g., \(\cos(2x)\) or \(\cos(7x)\))?
5. What is the significance of indeterminate forms in calculus?

**Tip:** Remember to check the form of a limit before applying L'Hopital's rule; it should be indeterminate like \(0/0\) or \(\infty/\infty\).

Learn how to solve the limit lim x→0 (x + cos(5x) - 1) / (-3x) using L'Hopital's rule with a detailed step-by-step solution. This problem involves limits and trigonometric functions, and is suitable for advanced high school students and beyond.

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