The limit in question is:

$\lim_{x \to 0} \frac{4x^2}{1 - e^x}$

Step 1: Evaluate the numerator as $x$ approaches 0.

The numerator $4x^2$ will approach 0 since $x^2$ tends to 0 when $x$ approaches 0.

$\lim_{x \to 0} 4x^2 = 0$

Step 2: Evaluate the denominator as $x$ approaches 0.

The denominator $1 - e^x$ can be approximated using the first-order Taylor expansion for $e^x$ around 0:

$e^x \approx 1 + x \text{ when } x \text{ is close to } 0$

So, the denominator becomes:

$1 - e^x \approx 1 - (1 + x) = -x$

Step 3: Combine the results.

Substituting the approximations into the limit:

$\lim_{x \to 0} \frac{4x^2}{-x} = \lim_{x \to 0} \frac{-4x^2}{x} = \lim_{x \to 0} -4x = 0$

Final Answer:

$\lim_{x \to 0} \frac{4x^2}{1 - e^x} = 0$

Would you like more details or have any questions?

Here are 5 related questions:

1. What happens to the limit if $x^2$ is replaced by $x^3$ in the numerator?
2. Can you derive the Taylor series for $e^x$ up to the second order?
3. How does the limit change if the denominator is $1 - e^{-x}$?
4. What is the limit $\lim_{x \to 0} \frac{x^2}{e^x - 1}$?
5. How does the approximation of $e^x$ affect the limit at higher values of $x$?

Tip: When dealing with limits that involve exponential functions, Taylor expansions can simplify the expressions and make it easier to evaluate the limit.