Question: lim x\to\infty \left( \frac{4^x + 9^x}{x} \right)^{\frac{1}{x}}
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 parentheses:**
As $$x \to \infty$$, compare $$4^x$$ and $$9^x$$. Since $$9^x$$ grows faster than $$4^x$$, $$9^x$$ dominates.
3. **Approximate the numerator:**
$$4^x + 9^x \approx 9^x$$ for large $$x$$.
4. **Rewrite the expression:**
$$\left( \frac{4^x + 9^x}{x} \right)^{\frac{1}{x}} \approx \left( \frac{9^x}{x} \right)^{\frac{1}{x}} = \left( 9^x \cdot \frac{1}{x} \right)^{\frac{1}{x}} = \left(9^x\right)^{\frac{1}{x}} \cdot \left( \frac{1}{x} \right)^{\frac{1}{x}}.$$
5. **Simplify each factor:**
$$\left(9^x\right)^{\frac{1}{x}} = 9^{x \cdot \frac{1}{x}} = 9,$$
and
$$\left( \frac{1}{x} \right)^{\frac{1}{x}} = e^{\frac{1}{x} \ln \frac{1}{x}} = e^{-\frac{\ln x}{x}}.$$
6. **Evaluate the limit of the second factor:**
As $$x \to \infty$$, $$\frac{\ln x}{x} \to 0$$, so
$$e^{-\frac{\ln x}{x}} \to e^0 = 1.$$
7. **Combine results:**
$$\lim_{x \to \infty} \left( \frac{4^x + 9^x}{x} \right)^{\frac{1}{x}} = 9 \times 1 = 9.$$
**Final answer:** $$\boxed{9}.$$