Certainly! Let's find the limit:
[tex]\[
\lim_{x \rightarrow 0} \frac{1 - \cos(9x)}{x^2}
\][/tex]
### Step-by-Step Solution:
1. Identify the Indeterminate Form:
As [tex]\( x \)[/tex] approaches 0, the numerator [tex]\( 1 - \cos(9x) \)[/tex] approaches 0 because [tex]\( \cos(9x) \)[/tex] approaches 1. The denominator [tex]\( x^2 \)[/tex] also approaches 0. Thus, we have the [tex]\(\frac{0}{0}\)[/tex] indeterminate form, which suggests that L'Hôpital's Rule might be useful.
2. Apply L'Hôpital's Rule:
L'Hôpital's Rule states that for limits of the form [tex]\(\frac{0}{0}\)[/tex],
[tex]\[
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
\][/tex]
provided the limit on the right side exists.
3. Differentiate the Numerator and the Denominator:
Let [tex]\( f(x) = 1 - \cos(9x) \)[/tex] and [tex]\( g(x) = x^2 \)[/tex].
- The derivative of the numerator [tex]\( f(x) \)[/tex]:
[tex]\[
f'(x) = \frac{d}{dx} [1 - \cos(9x)] = 9 \sin(9x) \][/tex]
- The derivative of the denominator [tex]\( g(x) \)[/tex]:
[tex]\[
g'(x) = \frac{d}{dx} [x^2] = 2x
\][/tex]
4. Form the New Limit with the Derivatives:
[tex]\[
\lim_{x \rightarrow 0} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 0} \frac{9 \sin(9x)}{2x}
\][/tex]
5. Simplify the Limit Expression:
Observe that [tex]\( \lim_{x \to 0} \frac{\sin(9x)}{9x} = 1 \)[/tex], a standard trigonometric limit.
Therefore,
[tex]\[
\lim_{x \rightarrow 0} \frac{9 \sin(9x)}{2x} = 9 \cdot \lim_{x \rightarrow 0} \frac{\sin(9x)}{9x} \cdot \frac{9}{2}
\][/tex]
6. Evaluate the Limit:
[tex]\[
\lim_{x \rightarrow 0} \frac{9 \sin(9x)}{2x} = \frac{9}{2} \cdot 9 = \frac{81}{2}
\][/tex]
### Final Answer:
[tex]\[
\boxed{\frac{81}{2}}
\][/tex]
This completes our solution. The limit of [tex]\( \frac{1 - \cos(9x)}{x^2} \)[/tex] as [tex]\( x \)[/tex] approaches 0 is [tex]\(\frac{81}{2}\)[/tex].
