https://file.aiquickdraw.com/mathbot/file/1729970204953wijita3v.jpg

To find the limit:

\[
\lim_{{x \to 0}} \frac{x + \sin(5x)}{x + \cos(4x)}
\]

we’ll start by evaluating the expression at \( x = 0 \) directly. 

### Step 1: Direct Substitution
Substitute \( x = 0 \) in the numerator and denominator:

- Numerator: \( x + \sin(5x) = 0 + \sin(0) = 0 \)
- Denominator: \( x + \cos(4x) = 0 + \cos(0) = 1 \)

Since the denominator does not go to zero, we can conclude that the limit is:

\[
\frac{0}{1} = 0
\]

### Final Answer:
\[
\lim_{{x \to 0}} \frac{x + \sin(5x)}{x + \cos(4x)} = 0
\]

Would you like further explanation on this approach or other methods?

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Here are some related questions to deepen your understanding:

1. What if the denominator also went to zero? How would l'Hospital's Rule be applied in that case?
2. How does the behavior of trigonometric functions like \(\sin(x)\) and \(\cos(x)\) affect limits as \(x \to 0\)?
3. What happens if we change the limits to \(\infty\)? How would this affect the result?
4. How would this problem change if \(\sin(5x)\) were replaced by \(\tan(5x)\)?
5. How can we use Taylor series expansions to evaluate limits of trigonometric expressions near zero?

**Tip:** When finding limits involving trigonometric functions as \(x \to 0\), try direct substitution first to see if the result is indeterminate or straightforward.

Find the limit. Use l'Hospital's Rule where appropriate. lim (x→0) (x + sin(5x)) / (x + cos(4x))

Learn how to solve the limit problem lim (x→0) (x + sin(5x)) / (x + cos(4x)). This solution involves direct substitution and analysis of indeterminate forms using l'Hôpital's Rule. Suitable for students in Grades 11-12.

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