We are tasked with finding the limit:

$\lim_{x \to 0} \frac{\sin(5x)}{\sin(9x)}$

Step 1: Checking the form

Both $\sin(5x)$ and $\sin(9x)$ approach 0 as $x \to 0$, so we have the indeterminate form $\frac{0}{0}$, which means L'Hôpital's Rule can be applied.

Step 2: Applying L'Hôpital's Rule

L'Hôpital's Rule states that if $\lim_{x \to a} \frac{f(x)}{g(x)}$ results in the indeterminate form $\frac{0}{0}$, then:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

In this case, we will differentiate the numerator and the denominator:

• The derivative of $\sin(5x)$ is $5 \cos(5x)$.
• The derivative of $\sin(9x)$ is $9 \cos(9x)$.

So, the new limit becomes:

$\lim_{x \to 0} \frac{5 \cos(5x)}{9 \cos(9x)}$

Step 3: Evaluating the limit

As $x \to 0$, $\cos(5x) \to 1$ and $\cos(9x) \to 1$, so we now have:

$\frac{5 \cdot 1}{9 \cdot 1} = \frac{5}{9}$

Final Answer:

$\lim_{x \to 0} \frac{\sin(5x)}{\sin(9x)} = \frac{5}{9}$

Would you like more details or have any questions about the steps?

Here are some related questions:

1. What happens if you replace the sine functions with cosine functions?
2. How do we apply L'Hôpital's Rule to $\lim_{x \to 0} \frac{x^2}{\sin^2(x)}$?
3. What are other indeterminate forms that require L'Hôpital's Rule?
4. Can we use Taylor expansions to approximate the limit instead of L'Hôpital's Rule?
5. How would the limit change if the argument inside the sine function was a quadratic expression like $5x^2$?

Tip: Before using L'Hôpital’s Rule, always check if simpler algebraic manipulations or known limits can solve the problem without differentiation.