Let's solve the given limit problem:

$\lim_{x \to -\infty} e^{2x}(x^2 - 2).$

Step 1: Analyze the terms

• The exponential term $e^{2x}$ rapidly approaches 0 as $x \to -\infty$, because $2x \to -\infty$.
• The polynomial term $(x^2 - 2)$ grows indefinitely large in magnitude (because $x^2 \to \infty$ as $x \to -\infty$).

Thus, the product $e^{2x}(x^2 - 2)$ involves a rapidly vanishing exponential term multiplied by a rapidly growing polynomial.

Step 2: Simplify using the exponential dominance property

Exponential terms like $e^{2x}$ approach 0 faster than any polynomial term grows. Thus, regardless of how fast $(x^2 - 2)$ grows, the factor $e^{2x} \to 0$ dominates and forces the product to 0.

Step 3: Conclusion

$\lim_{x \to -\infty} e^{2x}(x^2 - 2) = 0.$

Let me know if you’d like further clarification or detailed steps!

Here are 5 related questions for practice:

1. $\lim_{x \to \infty} e^{-x}(x^2 + 3)$
2. $\lim_{x \to -\infty} e^{-3x}(x^3 - x)$
3. Evaluate $\lim_{x \to \infty} e^{2x}/(x^2 + 1)$
4. Prove that $e^{ax}$ dominates $x^n$ as $x \to -\infty$ (for $a < 0$).
5. Evaluate $\lim_{x \to -\infty} e^{x}(x - 5)$.

Tip: Exponential terms dominate over polynomials at extremes (positive or negative infinity).