The problem you provided is:

$\lim_{{x \to 0}} \frac{\tan(6x)}{\sin(2x)}$

To solve this, we can use limit properties for small angles. As $x \to 0$, we know that:

1. $\tan(6x) \approx 6x$
2. $\sin(2x) \approx 2x$

Substituting these approximations, we get:

$\lim_{{x \to 0}} \frac{\tan(6x)}{\sin(2x)} \approx \lim_{{x \to 0}} \frac{6x}{2x} = \lim_{{x \to 0}} 3 = 3$

Therefore, the answer is:

$\lim_{{x \to 0}} \frac{\tan(6x)}{\sin(2x)} = 3$

Would you like more details on any part of the solution?

Related Questions

1. How would this solution change if the limit approached infinity instead?
2. What if we replaced $\tan(6x)$ with $\sin(6x)$? How would that affect the solution?
3. Can this method be used for limits involving $\tan(kx)$ and $\sin(kx)$ as $x \to \pi$?
4. What if we used L'Hôpital's Rule instead? Would we get the same result?
5. How would this problem look in polar coordinates?

Tip

When evaluating trigonometric limits as $x \to 0$, substituting approximations like $\sin(x) \approx x$ and $\tan(x) \approx x$ can simplify the process considerably.