To find the limit of the function $\lim_{{x \to 0}} \frac{\sin(2x)}{\sin(4x)}$ using the double angle identity, let's start by applying the double angle identities for sine:

$\sin(2x) = 2\sin(x)\cos(x)$ $\sin(4x) = 2\sin(2x)\cos(2x)$

Now, substitute $\sin(2x)$ in the denominator:

$\frac{\sin(2x)}{\sin(4x)} = \frac{2\sin(x)\cos(x)}{2\sin(2x)\cos(2x)}$

Simplifying by canceling the common factor of 2:

$\frac{\sin(2x)}{\sin(4x)} = \frac{\sin(x)\cos(x)}{\sin(2x)\cos(2x)}$

Now, substitute the expression $\sin(2x)$ again:

$\frac{\sin(x)\cos(x)}{2\sin(x)\cos(x)\cos(2x)} = \frac{1}{2\cos(2x)}$

Taking the limit as $x \to 0$:

$\lim_{{x \to 0}} \frac{1}{2\cos(2x)} = \frac{1}{2\cos(0)} = \frac{1}{2 \times 1} = \frac{1}{2}$

So, the limit is:

$\lim_{{x \to 0}} \frac{\sin(2x)}{\sin(4x)} = \frac{1}{2}$

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the limit of $\lim_{{x \to 0}} \frac{\sin(3x)}{\sin(5x)}$?
2. How does L'Hôpital's Rule apply to trigonometric limits?
3. What is the Taylor series expansion for $\sin(x)$ near $x = 0$?
4. How do you evaluate limits involving sine and cosine functions using their small-angle approximations?
5. What is the limit $\lim_{{x \to 0}} \frac{1 - \cos(x)}{x^2}$ and how is it derived?

Tip: For small angles, the sine function can be approximated by $\sin(x) \approx x$, which is useful in limit problems.