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?

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.