The given expression is:

$\lim_{x \to 0} \frac{x^2 - 3\sin(x)}{x}$

Let's solve this limit step by step:

Step 1: Evaluate directly at $x = 0$

Substitute $x = 0$ into the function:

$\frac{0^2 - 3\sin(0)}{0} = \frac{0 - 0}{0} = \frac{0}{0}$

This results in an indeterminate form $\frac{0}{0}$, so we need to apply L'Hôpital's Rule or simplify the expression further.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule states that for indeterminate forms like $\frac{0}{0}$, we can differentiate the numerator and the denominator and then take the limit.

• The derivative of the numerator $x^2 - 3\sin(x)$ is: $\frac{d}{dx} (x^2 - 3\sin(x)) = 2x - 3\cos(x)$

• The derivative of the denominator $x$ is: $\frac{d}{dx} x = 1$

Now the limit becomes: $\lim_{x \to 0} \frac{2x - 3\cos(x)}{1} = \lim_{x \to 0} (2x - 3\cos(x))$

Step 3: Evaluate the new limit

Substitute $x = 0$ into the new expression: $2(0) - 3\cos(0) = 0 - 3(1) = -3$

Final Answer:

The limit is:

$\boxed{-3}$

Would you like a detailed explanation on any part of this process?

Here are 5 related questions to expand on this concept:

1. How does L'Hôpital's Rule work for indeterminate forms other than $\frac{0}{0}$?
2. Can L'Hôpital's Rule always be applied when faced with indeterminate forms?
3. What is the significance of the $\sin(x)$ and $\cos(x)$ behavior near zero?
4. How would this limit change if there were a higher power of $x$ in the denominator?
5. What are some common tricks for evaluating limits without using L'Hôpital's Rule?

Tip: Always check if a limit leads to an indeterminate form before deciding to use L'Hôpital's Rule.