We are tasked with finding the limit:

$\lim_{x \to -\infty} e^x \cdot x^2$

Step-by-Step Solution:

1. Behavior of $e^x$ as $x \to -\infty$:

   • $e^x$ decays exponentially to $0$ as $x \to -\infty$.

2. Behavior of $x^2$ as $x \to -\infty$:

   • $x^2$ grows quadratically, but this growth is much slower compared to the decay of $e^x$.

3. Combined Analysis:

   • The term $e^x \cdot x^2$ combines the rapid decay of $e^x$ and the slower growth of $x^2$.
   • As $x \to -\infty$, $e^x$ dominates $x^2$, forcing the product to $0$.

Conclusion:

$\lim_{x \to -\infty} e^x \cdot x^2 = 0$

Let me know if you want further clarification or more detailed steps!

Related Questions to Explore:

1. What happens to $\lim_{x \to -\infty} e^x \cdot x$?
2. How does the exponential function compare to polynomial growth in limits?
3. Can the growth of $x^n$ (for $n > 2$) ever overpower $e^x$?
4. What is $\lim_{x \to +\infty} e^x \cdot x^2$?
5. What is the general rule for $e^x \cdot f(x)$ as $x \to -\infty$, where $f(x)$ is a polynomial?

Tip: In limits involving exponential decay, the exponential function typically dominates over polynomial or algebraic terms as (x \to -\infty