One more Limit

$\lim_{n \to \infty} \sqrt[n]{2^n + 3^n} = \lim_{n \to \infty} 3 \cdot \sqrt[n]{\left( \frac{2}{3} \right)^n + 1} = 3.$

Relevant wiki: Stolz–Cesàro theorem

$\begin{aligned} L & = \lim_{n \to \infty} \sqrt[n]{2^n+3^n} \\ & = \lim_{n \to \infty} \exp \left(\ln (2^n+3^n)^\frac 1n\right) & \small \color{#3D99F6} \text{where }\exp(x) = e^x \\ & = \exp \left(\lim_{n \to \infty} \frac {\ln (2^n+3^n)}n \right) & \small \color{#3D99F6} \text{By Stolz-Cesàro theorem} \\ & = \exp \left(\lim_{n \to \infty} \frac {\ln (2^{n+1}+3^{n+1}) - \ln (2^n+3^n)}{n+1-n} \right) \\ & = \exp \left(\lim_{n \to \infty} \ln \frac {2^{n+1}+3^{n+1}}{2^n+3^n} \right) \\ & = \exp \left(\lim_{n \to \infty} \ln \frac {\frac {2^{n+1}}{3^n} + \frac {3^{n+1}}{3^n}}{\frac {2^n}{3^n} + \frac {3^n}{3^n}} \right) \\ & = \exp \left(\ln \frac {0+3}{0+1} \right) = \exp (\ln 3) = \boxed{3} \end{aligned}$