1. **State the problem:** Find the limit $$\lim_{x \to \infty} \frac{8x}{e^{9x} + 1}$$ 2. **Check the form of the limit:** As $x \to \infty$, numerator $8x \to \infty$ and denominator $e^{9x} + 1 \to \infty$, so the limit is of the form $\frac{\infty}{\infty}$. This allows us to apply L'Hospital's Rule. 3. **Apply L'Hospital's Rule:** Differentiate numerator and denominator separately with respect to $x$: Numerator derivative: $\frac{d}{dx}(8x) = 8$ Denominator derivative: $\frac{d}{dx}(e^{9x} + 1) = 9 e^{9x}$ 4. **Rewrite the limit using these derivatives:** $$\lim_{x \to \infty} \frac{8}{9 e^{9x}}$$ 5. **Evaluate the limit:** As $x \to \infty$, $e^{9x} \to \infty$, so the denominator $9 e^{9x} \to \infty$, thus $$\frac{8}{9 e^{9x}} \to 0$$ 6. **Conclusion:** $$\lim_{x \to \infty} \frac{8x}{e^{9x} + 1} = 0$$