To find the limit [tex]\(\lim_{x \rightarrow 0} \frac{1 - \cos(2x)}{x^2}\)[/tex], we can follow these steps:
1. Recognize the Indeterminate Form: First, check the form of the expression as [tex]\(x\)[/tex] approaches 0. Both the numerator [tex]\(1 - \cos(2x)\)[/tex] and the denominator [tex]\(x^2\)[/tex] approach 0 as [tex]\(x\)[/tex] approaches 0, making this an indeterminate form of the type [tex]\(\frac{0}{0}\)[/tex].
2. Apply L'Hôpital's Rule: When faced with an indeterminate form of type [tex]\(\frac{0}{0}\)[/tex], one method to evaluate the limit is L'Hôpital's Rule. L'Hôpital's Rule states that:
[tex]\[
\lim_{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x \rightarrow 0} \frac{f'(x)}{g'(x)},
\][/tex]
provided the limit on the right exists. Here, [tex]\(f(x) = 1 - \cos(2x)\)[/tex] and [tex]\(g(x) = x^2\)[/tex].
3. Differentiate the Numerator and Denominator:
- The derivative of the numerator [tex]\(f(x) = 1 - \cos(2x)\)[/tex] is [tex]\(f'(x) = 2 \sin(2x)\)[/tex] using the chain rule.
- The derivative of the denominator [tex]\(g(x) = x^2\)[/tex] is [tex]\(g'(x) = 2x\)[/tex].
4. Form the New Limit: Substitute the derivatives back into the limit expression:
[tex]\[
\lim_{x \rightarrow 0} \frac{1 - \cos(2x)}{x^2} = \lim_{x \rightarrow 0} \frac{2 \sin(2x)}{2x}.
\][/tex]
5. Simplify the Expression:
[tex]\[
\lim_{x \rightarrow 0} \frac{2 \sin(2x)}{2x} = \lim_{x \rightarrow 0} \frac{\sin(2x)}{x}.
\][/tex]
6. Adjust the Argument of the Sine Function:
To simplify further, notice that:
[tex]\[
\frac{\sin(2x)}{x} = \frac{\sin(2x)}{2x} \cdot 2.
\][/tex]
Recognize that as [tex]\(x\)[/tex] approaches 0, [tex]\(\frac{\sin(2x)}{2x}\)[/tex] approaches 1 due to the standard limit [tex]\(\lim_{u \rightarrow 0} \frac{\sin(u)}{u} = 1\)[/tex].
7. Evaluate the Limit:
[tex]\[
\lim_{x \rightarrow 0} \frac{\sin(2x)}{x} = \lim_{x \rightarrow 0} \left( \frac{\sin(2x)}{2x} \cdot 2 \right) = 1 \cdot 2 = 2.
\][/tex]
Therefore, the solution to the limit is:
[tex]\[
\lim_{x \rightarrow 0} \frac{1 - \cos(2x)}{x^2} = 2.
\][/tex]
