Limit problem: $\lim\limits_{n\to ∞} \sqrt[n]{7^n+3^n}$ $$\lim_{n\to ∞} \sqrt[n]{7^n+3^n}$$

The answer should be $7$, but I have no idea how to get there.
Factor out the biggest term:
$$L = \lim_{n \to +\infty} \sqrt[n]{7^n\left(1 + \left(\frac37\right)^n\right)} = \lim_{n \to +\infty} 7\sqrt[n]{1 + \left(\frac37\right)^n} = 7,$$
since $3/7 &lt; 1$ and $(3/7)^n \to 0$.

You can easily generalize the above to prove:
$$\lim_{n \to +\infty} \left(a_0^n + \cdots + a_k^n\right)^{1/n} = \max\{a_0, \ldots, a_k\},$$
with $a_0,\ldots,a_k \in \mathbb R^+$.
$$\lim_{n\to +\infty}\sqrt[n]{7^n+3^n}$$
$$=\lim_{n\to +\infty}\left(7^n+3^n\right)^{1/n}$$
$$=\lim_{n\to +\infty}7\left(1+\frac{3^n}{7^n}\right)^{1/n}$$
$$=7\lim_{n\to +\infty}\left(1+\left(\frac37\right)^n\right)^{1/n}$$

since, $\frac 37&lt;1\implies $ $\lim_{n\to +\infty}\left(\frac 37\right)^n=0$, 
$$=7\left(1+0\right)^{0}=\color{red}{7}$$