The given image contains the following mathematical expression:

$\lim_{x \to 0} \frac{1 - \cos(x)}{x}$

To evaluate this limit, we can apply L'Hopital's Rule, since directly substituting $x = 0$ results in an indeterminate form $\frac{0}{0}$.

Step-by-Step Solution:

1. Apply L'Hopital's Rule: This rule is used for limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. It states that:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

if the limit on the right exists.

In this case, both the numerator $1 - \cos(x)$ and the denominator $x$ approach 0 as $x \to 0$. So we can apply L'Hopital's Rule by differentiating the numerator and the denominator:

• Derivative of the numerator: $\frac{d}{dx}[1 - \cos(x)] = \sin(x)$
• Derivative of the denominator: $\frac{d}{dx}[x] = 1$

2. Rewrite the limit:

$\lim_{x \to 0} \frac{1 - \cos(x)}{x} = \lim_{x \to 0} \frac{\sin(x)}{1}$

3. Evaluate the limit:

As $x \to 0$, $\sin(x) \to 0$. Hence:

$\lim_{x \to 0} \sin(x) = 0$

So, the value of the original limit is:

$\boxed{0}$

Would you like more details, or do you have any specific questions?

Here are five questions to extend this topic:

1. What is L'Hopital's Rule, and when can it be applied?
2. How would the limit change if the denominator were $x^2$ instead of $x$?
3. Can you explain why the function $\frac{1 - \cos(x)}{x}$ approaches 0 as $x \to 0$?
4. How can you approximate $\cos(x)$ for small values of $x$ using a Taylor series?
5. What are some alternative methods to evaluate trigonometric limits without L'Hopital's Rule?

Tip: Always check if the limit leads to an indeterminate form like $\frac{0}{0}$ before applying L'Hopital's Rule!