To evaluate the limit of the function

$\lim_{{x \to 0}} \frac{1 - \cos(4x)}{1 - \cos(2x)}$

let’s begin by checking for any indeterminate form. As $x \to 0$, both the numerator and denominator approach 0, which suggests that we may be dealing with a $\frac{0}{0}$ indeterminate form, and L'Hôpital's Rule could be applicable.

Step 1: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if the limit results in $\frac{0}{0}$, we can differentiate the numerator and denominator separately, and then take the limit of the resulting function. So we differentiate the numerator and denominator with respect to $x$.

Numerator:

$\frac{d}{dx}[1 - \cos(4x)] = 0 + 4\sin(4x)$

Denominator:

$\frac{d}{dx}[1 - \cos(2x)] = 0 + 2\sin(2x)$

Now, applying L'Hôpital's Rule, we get:

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

Step 2: Simplify the expression

We can simplify the new expression:

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

Step 3: Evaluate the limit

Now, using the fact that $\lim_{{x \to 0}} \frac{\sin(kx)}{kx} = 1$ for any constant $k$, we can simplify the limit further.

We recognize that: $\frac{\sin(4x)}{4x} \to 1 \quad \text{and} \quad \frac{\sin(2x)}{2x} \to 1 \quad \text{as} \quad x \to 0$

Thus, the limit becomes:

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

Final Answer:

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

Would you like a more detailed explanation of any of these steps? Or have any other questions?

Related Questions:

1. How does L'Hôpital's Rule work for limits involving trigonometric functions?
2. What are some other methods to handle indeterminate forms like $\frac{0}{0}$?
3. How do the small-angle approximations of $\sin(x) \approx x$ and $\cos(x) \approx 1 - \frac{x^2}{2}$ apply here?
4. Can you apply the same method to other trigonometric limits like $\lim_{{x \to 0}} \frac{1 - \cos(x)}{x^2}$?
5. How do Taylor series expansions help in evaluating limits involving trigonometric functions?

Tip:

When facing indeterminate forms, it’s always helpful to check if L'Hôpital's Rule can be applied before resorting to complex series expansions or other methods.