Limit Exponential 94E08C

1. **State the problem:** We want to find the limit $$\lim_{x \to \infty}\left(\frac{4^x + 9^x}{x}\right)^{\frac{1}{x}}.$$\n\n2. **Analyze the expression inside the limit:** The expression inside the parentheses is $$\frac{4^x + 9^x}{x}.$$ As $x \to \infty$, $9^x$ grows faster than $4^x$, so $9^x$ dominates $4^x$.\n\n3. **Approximate the dominant term:** We approximate $$4^x + 9^x \approx 9^x$$ for large $x$. So the expression becomes approximately $$\frac{9^x}{x}.$$\n\n4. **Rewrite the limit using this approximation:** $$\lim_{x \to \infty} \left(\frac{9^x}{x}\right)^{\frac{1}{x}}.$$\n\n5. **Simplify the expression inside the limit:** We can write $$\left(\frac{9^x}{x}\right)^{\frac{1}{x}} = \frac{(9^x)^{\frac{1}{x}}}{x^{\frac{1}{x}}} = \frac{9}{x^{\frac{1}{x}}}.$$\n\n6. **Evaluate the limit of the denominator:** As $x \to \infty$, $$x^{\frac{1}{x}} = e^{\frac{\ln x}{x}}.$$ Since $$\lim_{x \to \infty} \frac{\ln x}{x} = 0,$$ we have $$\lim_{x \to \infty} x^{\frac{1}{x}} = e^0 = 1.$$\n\n7. **Calculate the final limit:** Therefore, $$\lim_{x \to \infty} \frac{9}{x^{\frac{1}{x}}} = \frac{9}{1} = 9.$$\n\n**Final answer:** $$\boxed{9}.$$