Dominant Exponential 88A870

1. **Problem statement:** Evaluate $\lim_{x\to\infty}\left(4^x+9^x\right)^{1/x}$. 2. **Use the dominant-term rule:** For large $x$, the bigger exponential term grows fastest. Since $9^x$ grows faster than $4^x$, we factor out $9^x$. 3. **Rewrite the expression:** $$\left(4^x+9^x\right)^{1/x}=\left(9^x\left(\frac{4^x}{9^x}+1\right)\right)^{1/x}$$ 4. **Simplify the fraction inside:** $$\frac{4^x}{9^x}=\left(\frac{4}{9}\right)^x$$ So the expression becomes $$\left(9^x\left(\left(\frac{4}{9}\right)^x+1\right)\right)^{1/x}$$ 5. **Apply the $1/x$ power to each factor:** $$\left(9^x\right)^{1/x}\left(\left(\frac{4}{9}\right)^x+1\right)^{1/x}$$ 6. **Evaluate each part:** $$\left(9^x\right)^{1/x}=9$$ and since $\left(\frac{4}{9}\right)^x\to 0$, we have $$\left(\left(\frac{4}{9}\right)^x+1\right)^{1/x}\to 1$$ 7. **Final answer:** $$\lim_{x\to\infty}\left(4^x+9^x\right)^{1/x}=9$$