To solve the limit $\lim_{x \to 0} \frac{\sin(7x) + \tan(3x) - \sin(5x)}{\tan(9x) - \tan(3x) - \sin(x)}$ without substitution, we will use the fact that for small values of $x$, $\sin(x) \approx x$ and $\tan(x) \approx x$. This allows us to approximate each trigonometric function by its linear approximation near 0.

Let's approximate the functions:

$\sin(7x) \approx 7x, \quad \sin(5x) \approx 5x, \quad \sin(x) \approx x$ $\tan(3x) \approx 3x, \quad \tan(9x) \approx 9x$

Substituting these approximations into the limit:

$\lim_{x \to 0} \frac{7x + 3x - 5x}{9x - 3x - x}$

Simplifying the expressions:

$\lim_{x \to 0} \frac{(7x + 3x - 5x)}{(9x - 3x - x)} = \lim_{x \to 0} \frac{5x}{5x} = \lim_{x \to 0} 1 = 1$

Thus, the limit is $\boxed{1}$.

Would you like more details or have any questions about this solution? Here are some related questions for further exploration:

1. What is the limit of $\lim_{x \to 0} \frac{\sin(2x) - \sin(x)}{\tan(4x) - \tan(2x)}$?
2. How do small angle approximations help in evaluating trigonometric limits?
3. Can the same technique be used for limits involving $\cos(x)$?
4. How does the Taylor series expansion of trigonometric functions relate to this limit?
5. What happens if higher-order approximations are used for $\sin(x)$ and $\tan(x)$?

Tip: When working with limits of trigonometric functions as $x$ approaches 0, small angle approximations ($\sin(x) \approx x$ and $\tan(x) \approx x$) are often very useful for simplifying the expression.