To evaluate the limit
[tex]\[
\lim_{{x \to 0}} \frac{4x^2}{1 - \cos(2x)},
\][/tex]
let's go through the steps required to find this limit step-by-step.
1. Understanding the Problem: We need to find the limit of the given function as [tex]\( x \)[/tex] approaches 0. The denominator contains a trigonometric function, specifically [tex]\(\cos(2x)\)[/tex]. Cosine values close to 0 help hint towards using trigonometric identities or series expansions.
2. Trigonometric Identity: Recall that [tex]\(\cos(2x) = 1 - 2 \sin^2(x)\)[/tex]. This identity will be useful in simplifying the expression.
3. Simplify the Expression:
[tex]\[
1 - \cos(2x) = 1 - (1 - 2\sin^2(x)) = 2\sin^2(x).
\][/tex]
Substitute this into the original limit expression:
[tex]\[
\frac{4x^2}{1 - \cos(2x)} = \frac{4x^2}{2\sin^2(x)}.
\][/tex]
4. Further Simplification: Simplify the fraction:
[tex]\[
\frac{4x^2}{2\sin^2(x)} = \frac{2x^2}{\sin^2(x)}.
\][/tex]
5. Use Limit Properties: To evaluate this limit as [tex]\( x \to 0 \)[/tex], recall the standard limit:
[tex]\[
\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1.
\][/tex]
Therefore:
[tex]\[
\left(\frac{\sin(x)}{x}\right)^2 \to 1^2 = 1 \text{ as } x \to 0.
\][/tex]
6. Final Step: Rewrite [tex]\( \sin^2(x) \)[/tex] in the denominator using the limit:
[tex]\[
\frac{2x^2}{\sin^2(x)} = 2 \left(\frac{x}{\sin(x)}\right)^2.
\][/tex]
As [tex]\( x \to 0 \)[/tex]:
[tex]\[
\left(\frac{x}{\sin(x)}\right)^2 \to 1.
\][/tex]
So, the whole expression approaches:
[tex]\[
2 \times 1 = 2.
\][/tex]
Therefore,
[tex]\[
\lim_{{x \to 0}} \frac{4x^2}{1 - \cos(2x)} = 2.
\][/tex]
