Exponential Limit 2D976A

1. **Problem:** Evaluate $\displaystyle \lim_{x\to\infty}\left(\frac{4^x+9^x}{x}\right)^{1/x}$. 2. Factor out the dominant exponential term, $9^x$, inside the fraction: $$\left(\frac{4^x+9^x}{x}\right)^{1/x}=9\left(\frac{1+(4/9)^x}{x}\right)^{1/x}.$$ 3. Use logarithms to evaluate the limit. For the logarithm of the original expression, we get $$\frac{1}{x}\ln\left(\frac{4^x+9^x}{x}\right)=\ln 9+\frac{\ln\left(1+(4/9)^x\right)-\ln x}{x}.$$ 4. As $x\to\infty$, $(4/9)^x\to 0$, so $\ln\left(1+(4/9)^x\right)\to 0$. Also, $\frac{\ln x}{x}\to 0$. Therefore, the logarithm tends to $\ln 9$. 5. Exponentiating gives the final answer: $$\boxed{9}$$