The limit presented in the image is:

$\lim_{x \to 0} \frac{1 - \cos^2(2x)}{(2x)^2}$

We can simplify and solve this limit step-by-step.

Step 1: Use a trigonometric identity

We know the identity $1 - \cos^2(\theta) = \sin^2(\theta)$. Applying this to the numerator:

$1 - \cos^2(2x) = \sin^2(2x)$

So the limit becomes:

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

Step 2: Simplify the expression

We now rewrite the expression as:

$\lim_{x \to 0} \left( \frac{\sin(2x)}{2x} \right)^2$

Step 3: Use the standard limit

We know from standard calculus that:

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

So, applying this to $\frac{\sin(2x)}{2x}$:

$\lim_{x \to 0} \left( \frac{\sin(2x)}{2x} \right)^2 = 1^2 = 1$

Final Answer:

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

Would you like any further details or explanations on this solution?

Here are 5 related questions to expand your understanding:

1. How does the limit of $\frac{\sin(x)}{x}$ help in solving other trigonometric limits?
2. What other trigonometric identities can be applied to simplify limits?
3. How would the limit change if we were using a different trigonometric function, such as $\tan(x)$?
4. What are common standard limits used in calculus, and why are they important?
5. How do you handle indeterminate forms like $\frac{0}{0}$ in limits?

Tip: Always check for trigonometric identities when dealing with limits involving trigonometric functions—they can simplify the problem significantly!