The expression in the image is:

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

To solve this limit, we can start by evaluating the limit directly:

1. Substitute $x = 0$ into the expression:

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

   The result is an indeterminate form $\frac{0}{0}$, so we need to apply L'Hôpital's rule.

2. Apply L'Hôpital's rule:

   L'Hôpital's rule states that if the limit results in an indeterminate form like $\frac{0}{0}$, we can differentiate the numerator and the denominator separately and then take the limit:

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

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

   Now, the limit becomes: $\lim_{x \to 0} \frac{6x}{-e^x}$

3. Evaluate the new limit:

   Substitute $x = 0$ into the new expression: $\frac{6(0)}{-e^0} = \frac{0}{-1} = 0$

So, the limit is:

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

Would you like any further details or explanations? Here are some additional questions to consider:

1. How is L'Hôpital's rule applied when the limit results in $\frac{\infty}{\infty}$?
2. What are some alternative methods to solve limits without using L'Hôpital's rule?
3. How do you differentiate $e^x$ and why is it the same as $e^x$?
4. What happens to the limit if the numerator had a higher power of $x$?
5. How can you verify the result using a series expansion for $e^x$?

Tip: L'Hôpital's rule is useful for handling limits that result in indeterminate forms, but it’s important to remember that the rule only applies when you have these specific forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.