To solve the limit [tex]\(\lim_{x \to 0} \frac{1 - \cos x}{x^2}\)[/tex], we can proceed with the following steps:
### Step 1: Recognize that direct substitution leads to an indeterminate form
If we directly substitute [tex]\(x = 0\)[/tex] into the expression [tex]\(\frac{1 - \cos x}{x^2}\)[/tex], we get:
[tex]\[
\frac{1 - \cos 0}{0^2} = \frac{1 - 1}{0^2} = \frac{0}{0}
\][/tex]
This is an indeterminate form. Hence, we need to use other methods to evaluate this limit.
### Step 2: Use the Taylor Series expansion for [tex]\(\cos x\)[/tex]
The Taylor series expansion for [tex]\(\cos x\)[/tex] around [tex]\(x = 0\)[/tex] is:
[tex]\[
\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots
\][/tex]
We can use this expansion to approximate [tex]\(\cos x\)[/tex] when [tex]\(x\)[/tex] is close to 0.
### Step 3: Substitute the Taylor Series into the limit expression
Using the Taylor Series expansion:
[tex]\[
1 - \cos x = 1 - \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \right) = \frac{x^2}{2} - \frac{x^4}{24} + \cdots
\][/tex]
### Step 4: Simplify the limit expression
Now, substitute this expansion into the limit:
[tex]\[
\frac{1 - \cos x}{x^2} = \frac{\frac{x^2}{2} - \frac{x^4}{24} + \cdots}{x^2}
\][/tex]
Divide each term in the numerator by [tex]\(x^2\)[/tex]:
[tex]\[
\frac{x^2 / 2 - x^4 / 24 + \cdots}{x^2} = \frac{x^2}{2x^2} - \frac{x^4}{24x^2} + \cdots = \frac{1}{2} - \frac{x^2}{24} + \cdots
\][/tex]
### Step 5: Take the limit as [tex]\(x \to 0\)[/tex]
As [tex]\(x\)[/tex] approaches 0, higher-order terms (like [tex]\(\frac{x^2}{24}\)[/tex], etc.) will approach 0. Therefore, we have:
[tex]\[
\lim_{x \to 0} \left( \frac{1}{2} - \frac{x^2}{24} + \cdots \right) = \frac{1}{2}
\][/tex]
### Conclusion
Thus, the limit is:
[tex]\[
\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}
\][/tex]
