Limit Exponential Root 755F49

1. **State the problem:** Find the limit $$\lim_{x \to \infty} \left(\frac{4^x + 9^x}{x}\right)^{\frac{1}{x}}$$ 2. **Analyze the expression inside the limit:** As $x \to \infty$, compare $4^x$ and $9^x$. Since $9^x$ grows faster than $4^x$, $9^x$ dominates $4^x$. So, $$4^x + 9^x \approx 9^x$$ 3. **Simplify the expression inside the parentheses:** $$\frac{4^x + 9^x}{x} \approx \frac{9^x}{x}$$ 4. **Rewrite the limit using this approximation:** $$\lim_{x \to \infty} \left(\frac{9^x}{x}\right)^{\frac{1}{x}}$$ 5. **Rewrite the expression inside the power:** $$\left(\frac{9^x}{x}\right)^{\frac{1}{x}} = \frac{(9^x)^{\frac{1}{x}}}{x^{\frac{1}{x}}} = \frac{9}{x^{\frac{1}{x}}}$$ 6. **Evaluate the limit of the denominator:** $$\lim_{x \to \infty} x^{\frac{1}{x}} = \lim_{x \to \infty} e^{\frac{\ln x}{x}}$$ Since $\lim_{x \to \infty} \frac{\ln x}{x} = 0$, we have $$\lim_{x \to \infty} x^{\frac{1}{x}} = e^0 = 1$$ 7. **Therefore, the limit becomes:** $$\lim_{x \to \infty} \frac{9}{x^{\frac{1}{x}}} = \frac{9}{1} = 9$$ **Final answer:** $$\boxed{9}$$