Certainly! The limit you've provided can be evaluated step by step using symbolic notation as follows:

$\lim_{x \to 0} \frac{3x^2}{1 - e^x}$

Step 1: Recognize the indeterminate form

First, we substitute $x = 0$ directly into the limit:

$\frac{3(0)^2}{1 - e^0} = \frac{0}{1 - 1} = \frac{0}{0}$

Since we have the indeterminate form $\frac{0}{0}$, we apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule requires us to differentiate the numerator and the denominator separately with respect to $x$.

Differentiate the numerator $3x^2$: $\frac{d}{dx}(3x^2) = 6x$

Differentiate the denominator $1 - e^x$: $\frac{d}{dx}(1 - e^x) = -e^x$

Now, we rewrite the limit using these derivatives: $\lim_{x \to 0} \frac{6x}{-e^x}$

Step 3: Evaluate the new limit

Substitute $x = 0$ into the expression:

$\frac{6(0)}{-e^0} = \frac{0}{-1} = 0$

Final Answer

Thus, the limit is:

$\lim_{x \to 0} \frac{3x^2}{1 - e^x} = 0$

If you have any further questions or need a deeper explanation, feel free to ask! Here are some related questions you might find useful:

1. How would you apply L'Hôpital's Rule if the limit were $\lim_{x \to 0} \frac{x^2}{\sin(x)}$?
2. What happens if you attempt to use the Taylor series expansion for $e^x$ in this limit?
3. Can you apply L'Hôpital's Rule multiple times for higher indeterminate forms?
4. What are the conditions under which L'Hôpital's Rule can be applied?
5. How does L'Hôpital's Rule compare to using a series expansion for limits near zero?

Tip: When evaluating limits, always check if direct substitution leads to an indeterminate form. If it does, L'Hôpital's Rule or series expansion methods can be valuable tools to find the limit.