The given limit is:

$\lim_{x \to 0} \frac{1 - \cos x}{x^2}$

To solve this limit, we can apply the known series expansion of $\cos x$ around $x = 0$ or apply L'Hopital's Rule since it is an indeterminate form of type $\frac{0}{0}$.

Step-by-step solution using L'Hopital's Rule:

1. First, we verify that the limit is of the indeterminate form $\frac{0}{0}$:

   • As $x \to 0$, $1 - \cos(0) = 1 - 1 = 0$, and $x^2 \to 0^2 = 0$.
   • Therefore, we have the indeterminate form $\frac{0}{0}$, and L'Hopital's Rule can be applied.

2. Differentiate the numerator and denominator with respect to $x$:

   • The derivative of $1 - \cos x$ is $\sin x$.
   • The derivative of $x^2$ is $2x$.

   So, applying L'Hopital's Rule, the limit becomes: $\lim_{x \to 0} \frac{\sin x}{2x}$

3. Now, evaluate the new limit. It is a standard limit: $\lim_{x \to 0} \frac{\sin x}{x} = 1$ Therefore, the limit becomes: $\lim_{x \to 0} \frac{\sin x}{2x} = \frac{1}{2}$

Thus, the value of the limit is:

$\boxed{\frac{1}{2}}$

Would you like further details or explanations on this? Here are some related questions:

1. What is the Taylor expansion of $\cos x$ and how can it be used to solve limits?
2. Can we solve this limit using another method like series expansion?
3. How does L'Hopital's Rule work for different indeterminate forms?
4. What other limits frequently lead to the use of L'Hopital's Rule?
5. Why is the limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ important?

Tip: When solving limits, always check for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, as they often suggest the use of L'Hopital's Rule.