Let's solve the given limit step-by-step.

Problem

We are solving the following: $\lim_{x \to 0} \left( \frac{\sin(7x)}{9x} + (x^2 + 1) \cdot \frac{\sin(9x)}{\sin(7x)} \right)$

Step 1: Simplify the first term

Using the small-angle approximation $\sin(kx) \approx kx$ as $x \to 0$, the first term becomes: $\frac{\sin(7x)}{9x} \to \frac{7x}{9x} = \frac{7}{9}.$

Step 2: Simplify the second term

The second term is $(x^2 + 1) \cdot \frac{\sin(9x)}{\sin(7x)}$.

(a) Simplify $\frac{\sin(9x)}{\sin(7x)}$:

Using $\sin(kx) \approx kx$, $\frac{\sin(9x)}{\sin(7x)} \to \frac{9x}{7x} = \frac{9}{7}.$

(b) Evaluate $x^2 + 1$ as $x \to 0$:

$x^2 + 1 \to 1.$

Thus, the second term becomes: $(x^2 + 1) \cdot \frac{\sin(9x)}{\sin(7x)} \to 1 \cdot \frac{9}{7} = \frac{9}{7}.$

Step 3: Combine the results

Adding the two terms: $\lim_{x \to 0} \left( \frac{\sin(7x)}{9x} + (x^2 + 1) \cdot \frac{\sin(9x)}{\sin(7x)} \right) = \frac{7}{9} + \frac{9}{7}.$

Step 4: Simplify the sum

To combine $\frac{7}{9}$ and $\frac{9}{7}$, find a common denominator: $\frac{7}{9} + \frac{9}{7} = \frac{49}{63} + \frac{81}{63} = \frac{130}{63}.$

Thus, the answer is: $\boxed{\frac{130}{63}}$

Would you like further clarification or a deeper breakdown?
Here are related questions to explore:

1. How does the small-angle approximation work for trigonometric functions?
2. Can you generalize the approach to similar trigonometric limits?
3. Why is it valid to assume $x^2 + 1 \approx 1$ as $x \to 0$?
4. How can L'Hôpital's Rule be used to solve such limits?
5. What are common mistakes when simplifying fractions in such problems?

Tip: Always verify trigonometric approximations when $x \to 0$ using Taylor series for better precision.