To evaluate the limit

$\lim_{x \to \infty} \sqrt{x} e^{-x/2},$

we proceed as follows:

Step 1: Rewrite the expression

The term $e^{-x/2}$ grows very small very quickly as $x \to \infty$, while $\sqrt{x}$ grows unbounded. Therefore, we expect the exponential decay to dominate. Let’s analyze it rigorously.

Step 2: Substitution for simplification

Rewrite the expression: $\sqrt{x} e^{-x/2} = \frac{\sqrt{x}}{e^{x/2}}.$ Now, focus on the dominant term in the denominator, $e^{x/2}$, which grows exponentially faster than $\sqrt{x}$.

Step 3: Apply logarithmic properties (optional)

To simplify the growth comparison, consider taking the natural logarithm of the numerator and denominator to analyze growth rates:

• $\ln(\sqrt{x}) = \frac{1}{2} \ln(x)$,
• while $\ln(e^{x/2}) = \frac{x}{2}$.

The term $\frac{x}{2}$ clearly dominates $\frac{1}{2} \ln(x)$ as $x \to \infty$, implying the entire fraction goes to 0.

Step 4: Rigorous confirmation using substitution

Let $u = \sqrt{x}$, so as $x \to \infty$, $u \to \infty$. Then: $\sqrt{x} e^{-x/2} = u e^{-u^2 / 2}.$ The term $e^{-u^2 / 2}$ approaches 0 extremely rapidly, much faster than $u$, confirming: $\lim_{x \to \infty} \sqrt{x} e^{-x/2} = 0.$

Final Answer:

$\lim_{x \to \infty} \sqrt{x} e^{-x/2} = 0.$

Would you like a step-by-step graphical explanation or derivation using another approach? Here are some additional questions you might find interesting:

1. How does exponential decay compare to polynomial growth for large values of $x$?
2. What is the effect of adding a coefficient to $x$ in the exponent of $e^{-x/2}$?
3. Can we generalize this limit to $\lim_{x \to \infty} x^a e^{-bx}$ for constants $a$ and $b$?
4. How does the rate of decay differ between $e^{-x}$ and $e^{-x^2}$?
5. What happens if $\sqrt{x}$ is replaced by $x^n$ where $n > 1$?

Tip: Exponential terms $e^{-x}$ dominate polynomial terms in limits as $x \to \infty$. Always analyze the growth/decay rates when comparing such terms!