Certainly! We can solve this limit problem without using L'Hôpital's Rule by leveraging the Taylor series expansion for $\sin(x)$ or by simplifying the expression using algebraic techniques.

Step 1: Use the Taylor Series Expansion

The Taylor series expansion of $\sin(x)$ around $x = 0$ is:

$\sin(x) = x - \frac{x^3}{6} + \text{higher-order terms}$

For small values of $x$, we can approximate $\sin(x)$ by its first term:

$\sin(x) \approx x$

Step 2: Simplify the Expression

Substitute this approximation into the limit:

$\lim_{{x \to 0}} \frac{5x^2 + 9x}{\sin(x)} \approx \lim_{{x \to 0}} \frac{5x^2 + 9x}{x}$

Now, factor the numerator:

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

Step 3: Simplify Further

Cancel the $x$ terms in the numerator and denominator (note that $x \neq 0$ for this cancellation):

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

Step 4: Evaluate the Limit

Now, directly substitute $x = 0$ into the simplified expression:

$5(0) + 9 = 9$

Thus, the limit is:

$\lim_{{x \to 0}} \frac{5x^2 + 9x}{\sin(x)} = 9$

This method avoids the use of L'Hôpital's Rule by using the approximation for $\sin(x)$ and straightforward algebra.

Do you have any more questions or would you like to explore other approaches?

Here are five related questions for further exploration:

1. How does the Taylor series for $\sin(x)$ help in finding limits?
2. Can you generalize the process for finding limits involving small-angle approximations?
3. What would the limit be if the numerator were $x^3$ instead of $5x^2 + 9x$?
4. How can we justify the cancellation of $x$ in the numerator and denominator?
5. How does this approach differ from using L'Hôpital's Rule in terms of understanding the behavior of the function?

Tip: Small-angle approximations like $\sin(x) \approx x$ are extremely useful for solving limits, especially when $x$ approaches zero.