Limit Exponential Acbf59

1. **State the problem:** Find the limit $$\lim_{x \to \infty} \left(\frac{4^x + 9^x}{x}\right)^{1/x}$$ 2. **Analyze the expression:** Inside the parentheses, we have a sum of exponential terms $4^x$ and $9^x$ divided by $x$, all raised to the power $1/x$. 3. **Dominant term:** As $x \to \infty$, $9^x$ grows faster than $4^x$, so $9^x$ dominates $4^x$. Thus, $$4^x + 9^x \sim 9^x$$ 4. **Rewrite the expression using the dominant term:** $$\left(\frac{4^x + 9^x}{x}\right)^{1/x} \approx \left(\frac{9^x}{x}\right)^{1/x}$$ 5. **Simplify the expression:** $$\left(\frac{9^x}{x}\right)^{1/x} = \frac{(9^x)^{1/x}}{x^{1/x}} = \frac{9^{x \cdot \frac{1}{x}}}{x^{1/x}} = \frac{9}{x^{1/x}}$$ 6. **Evaluate the limit of $x^{1/x}$ as $x \to \infty$:** Recall that $$x^{1/x} = e^{\frac{\ln x}{x}}$$ As $x \to \infty$, $\frac{\ln x}{x} \to 0$, so $$\lim_{x \to \infty} x^{1/x} = e^0 = 1$$ 7. **Calculate the final limit:** $$\lim_{x \to \infty} \frac{9}{x^{1/x}} = \frac{9}{1} = 9$$ **Final answer:** $$\boxed{9}$$