To solve the limit $\lim_{x \to 0} \frac{\tan(21x)}{\sin(30x)}$, we can proceed by using trigonometric approximations for small angles without applying L'Hôpital's rule, as requested.

Step-by-Step Solution

1. Rewrite Using Small Angle Approximations: For small angles (as $x \to 0$), we can use the approximations: $\tan(21x) \approx 21x$ and $\sin(30x) \approx 30x.$

2. Substitute the Approximations: Substitute these approximations into the expression: $\lim_{x \to 0} \frac{\tan(21x)}{\sin(30x)} \approx \lim_{x \to 0} \frac{21x}{30x}.$

3. Simplify the Expression: Since $x$ appears in both the numerator and the denominator, we can cancel $x$: $\lim_{x \to 0} \frac{21x}{30x} = \frac{21}{30}.$

4. Reduce the Fraction: Simplify $\frac{21}{30}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 3: $\frac{21}{30} = \frac{7}{10}.$

Conclusion

The limit is: $\lim_{x \to 0} \frac{\tan(21x)}{\sin(30x)} = \frac{7}{10}.$

Would you like a further explanation on any step, or do you have questions?

Related Questions

1. How does the small-angle approximation work in trigonometric limits?
2. Why is L'Hôpital's Rule not necessary for limits involving trigonometric functions?
3. What would change if we used $\cos$ instead of $\sin$ in the denominator?
4. How does the limit change if the angles are in degrees rather than radians?
5. Could this limit be solved using a Taylor series approach?

Tip

For trigonometric limits as $x \to 0$, always consider using small-angle approximations to simplify expressions without needing L'Hôpital's Rule.