Question: top-left: none
center: lim
x→∞
(4^x + 9^x)
────────────
x
)^(1/x)
Graph shape: none visible; only a mathematical limit expression with a large parenthesized fraction raised to the power 1/x.
1. **State the problem.**
We want to find
$$\lim_{x\to\infty}\left(\frac{4^x+9^x}{x}\right)^{1/x}.$$
2. **Identify the dominant term.**
As $x\to\infty$, the term $9^x$ grows faster than $4^x$, so the numerator is dominated by $9^x$.
Factor out $9^x$:
$$4^x+9^x=9^x\left(\frac{4^x}{9^x}+1\right)=9^x\left(\left(\frac{4}{9}\right)^x+1\right).$$
3. **Rewrite the expression.**
Substitute that into the limit:
$$\left(\frac{4^x+9^x}{x}\right)^{1/x}=\left(\frac{9^x\left(\left(\frac{4}{9}\right)^x+1\right)}{x}\right)^{1/x}.$$
Split the $1/x$ power across the factors:
$$=\left(9^x\right)^{1/x}\left(\left(\frac{4}{9}\right)^x+1\right)^{1/x}\left(\frac{1}{x}\right)^{1/x}.$$
4. **Simplify each factor.**
The first factor is
$$\left(9^x\right)^{1/x}=9.$$
The second factor satisfies
$$\left(\frac{4}{9}\right)^x\to 0,$$
so
$$\left(\left(\frac{4}{9}\right)^x+1\right)^{1/x}\to 1^{1/x}=1.$$
The third factor is
$$\left(\frac{1}{x}\right)^{1/x}.$$
Since $\frac{1}{x}\to 0^+$ and $\lim_{x\to\infty}x^{1/x}=1$, we get
$$\left(\frac{1}{x}\right)^{1/x}\to 1.$$
5. **Combine the limits.**
Therefore,
$$\lim_{x\to\infty}\left(\frac{4^x+9^x}{x}\right)^{1/x}=9\cdot 1\cdot 1=9.$$
6. **Final answer.**
$$\boxed{9}$$