Question: top-left: none
top-right: none
center: $\lim_{x\to\infty}\left(\frac{4^x+9^x}{x}\right)^{1/x}$
bottom-left: $x\to\infty$
bottom-right: none
**Problem:** Evaluate $\lim_{x\to\infty}\left(\frac{4^x+9^x}{x}\right)^{1/x}$.
1. Use the dominant-term rule: when $x\to\infty$, the larger exponential base dominates, so factor out $9^x$.
$$\left(\frac{4^x+9^x}{x}\right)^{1/x}=\left(\frac{9^x\left[\left(\frac{4}{9}\right)^x+1\right]}{x}\right)^{1/x}$$
2. Separate the factors using exponent rules.
$$=9\left(\left(\frac{4}{9}\right)^x+1\right)^{1/x}\left(\frac{1}{x}\right)^{1/x}$$
3. Since $0<\frac{4}{9}<1$, we have $\left(\frac{4}{9}\right)^x\to0$, so the middle factor approaches $1$.
$$\left(\left(\frac{4}{9}\right)^x+1\right)^{1/x}\to1$$
4. Also, $x^{1/x}\to1$, so its reciprocal approaches $1$.
$$\left(\frac{1}{x}\right)^{1/x}=\frac{1}{x^{1/x}}\to1$$
5. Multiply the limiting factors.
$$9\cdot1\cdot1=9$$
**Final answer:** $9$.