To evaluate the limit

$\lim_{x \to \infty} \frac{x^2}{e^{3x}}$

using L'Hôpital's Rule, we first examine if the rule applies. This is an $\frac{\infty}{\infty}$ form as $x \to \infty$, so L'Hôpital's Rule is appropriate.

Step 1: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator:

1. Numerator: The derivative of $x^2$ is $2x$.
2. Denominator: The derivative of $e^{3x}$ is $3e^{3x}$.

After applying L'Hôpital's Rule once, we have:

$\lim_{x \to \infty} \frac{2x}{3e^{3x}}$

Step 2: Apply L'Hôpital's Rule Again

The limit is still an $\frac{\infty}{\infty}$ form, so we apply L'Hôpital's Rule a second time.

1. Numerator: The derivative of $2x$ is $2$.
2. Denominator: The derivative of $3e^{3x}$ is $9e^{3x}$.

This gives us:

$\lim_{x \to \infty} \frac{2}{9e^{3x}}$

Step 3: Evaluate the Limit

As $x \to \infty$, $e^{3x}$ grows exponentially, making $9e^{3x} \to \infty$. Therefore, the fraction $\frac{2}{9e^{3x}}$ approaches 0.

Conclusion

The exact answer is:

$\lim_{x \to \infty} \frac{x^2}{e^{3x}} = 0$

Would you like more details on any part of this solution?

Related Questions:

1. How does L'Hôpital's Rule work for indeterminate forms?
2. What other types of indeterminate forms can L'Hôpital's Rule be used for?
3. Why does $e^{3x}$ grow faster than any polynomial as $x \to \infty$?
4. Can we use a different approach to evaluate this limit without L'Hôpital's Rule?
5. How does exponential growth compare to polynomial growth in limits?

Tip:

When using L'Hôpital's Rule, always confirm the form of the limit ($\infty/\infty$ or $0/0$) before applying the rule.