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.

Explore how to evaluate the limit of (4x^2)/(1 - e^x) as x approaches 0. Step-by-step solution using first-order Taylor series expansion for e^x. Learn about limits, exponential functions, and Taylor series. Suitable for college-level mathematics.

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