To solve the limit problem $\lim_{x \to 0} \frac{\sin(7x) + \tan(3x) - \sin(5x)}{\tan(9x) - \tan(3x) - \sin(x)}$, we can use the small angle approximations for trigonometric functions:

For $x$ close to $0$: $\sin(kx) \approx kx \quad \text{and} \quad \tan(kx) \approx kx$

Step-by-Step Solution:

1. Apply the small angle approximations: $\sin(7x) \approx 7x, \quad \tan(3x) \approx 3x, \quad \sin(5x) \approx 5x$ $\tan(9x) \approx 9x, \quad \tan(3x) \approx 3x, \quad \sin(x) \approx x$

2. Substitute these approximations into the limit: $\lim_{x \to 0} \frac{7x + 3x - 5x}{9x - 3x - x}$

3. Simplify the expression: $\lim_{x \to 0} \frac{(7x + 3x - 5x)}{(9x - 3x - x)} = \lim_{x \to 0} \frac{5x}{5x}$

4. Simplify further: $\lim_{x \to 0} 1 = 1$

Final Answer:

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

Would you like more details, or do you have any questions?

Related Questions:

1. How would the limit change if the expression involved $\cos(x)$ instead of $\sin(x)$?
2. What happens if we change the coefficients of $x$ in the sine and tangent functions?
3. Can we apply L'Hôpital's Rule to solve this limit instead of small-angle approximations?
4. How does the limit change if $x$ approaches a different value (e.g., $\pi/2$)?
5. What are the differences in solving limits involving trigonometric functions compared to polynomial functions?

Tip:

For small values of $x$, always consider using the small-angle approximations for trigonometric functions like $\sin(x)$ and $\tan(x)$ to simplify the limit calculation.