The given expression is:

$\lim_{x \to 0} \frac{3(1 - \cos x)}{x}$

We can solve this limit using a standard trigonometric limit or L'Hôpital's Rule. Here are both approaches:

Using L'Hôpital's Rule:

Since direct substitution of $x = 0$ leads to the indeterminate form $\frac{0}{0}$, we apply L'Hôpital's Rule, which requires us to differentiate the numerator and denominator with respect to $x$.

• Numerator: $3(1 - \cos x)$. The derivative is $3 \cdot (\sin x)$.
• Denominator: $x$. The derivative is $1$.

So, the limit becomes:

$\lim_{x \to 0} \frac{3\sin x}{1} = 3\sin(0) = 0$

Thus, the limit is:

$\boxed{0}$

Would you like more details, or do you have any other questions? Here are 5 related questions to further your understanding:

1. How do we know when to apply L'Hôpital's Rule?
2. Can we solve this limit without using L'Hôpital's Rule?
3. What other indeterminate forms can be solved using L'Hôpital's Rule?
4. How does the small-angle approximation $\cos x \approx 1 - \frac{x^2}{2}$ help in solving limits like this?
5. What are other common limits involving trigonometric functions?

Tip: When evaluating limits of trigonometric functions at zero, recognizing common indeterminate forms and knowing key limit results can simplify your work considerably.