To solve the given problem, we analyze the function $f(x) = 2xe^{2x}$ and compute the required limits as $x \to -\infty$ and $x \to \infty$.

(a) Find $\lim_{x \to -\infty} f(x)$ and $\lim_{x \to \infty} f(x)$:

1. As $x \to -\infty$:

   • Consider $e^{2x}$, which decays to 0 exponentially as $x \to -\infty$.
   • The term $2x$ grows negatively, but the product $2x \cdot e^{2x}$ is dominated by the exponential decay of $e^{2x}$. Therefore: $\lim_{x \to -\infty} 2xe^{2x} = 0.$

2. As $x \to \infty$:

   • $e^{2x}$ grows exponentially, while $x$ grows linearly.
   • The product $2x \cdot e^{2x}$ is dominated by the exponential growth of $e^{2x}$. Therefore: $\lim_{x \to \infty} 2xe^{2x} = \infty.$

Final Answer:

$\lim_{x \to -\infty} f(x) = 0, \quad \lim_{x \to \infty} f(x) = \infty.$

Would you like a step-by-step explanation or further analysis?

Related Questions:

1. How does the growth rate of $e^{2x}$ compare to $x^n$ for any integer $n$?
2. What happens to $f(x) = 2xe^{2x}$ if the coefficient of $x$ is negative instead of positive?
3. How can L'Hôpital's Rule help confirm the behavior of $f(x)$ as $x \to \infty$?
4. How does the rate of decay of $e^{2x}$ influence the limit as $x \to -\infty$?
5. What are some practical applications of analyzing exponential functions like $f(x) = 2xe^{2x}$?

Tip: When evaluating limits involving exponential terms, always compare growth rates (e.g., linear vs. exponential). Exponentials dominate in the long term.