Solution: Without loss of generality, let us assume that $\alpha \geq \beta $. We will show that \[ \lim_{n \to \infty} \sqrt[n]{\alpha ^n + \beta ^n} = \alpha . \] Note that \begin{align*} \lim_{n \to \infty} \sqrt[n]{\alpha^n + \beta ^n} & = \lim_{n \to \infty} \sqrt[n]{\alpha ^n \left( 1 + \frac{\beta ^n}{\alpha ^n} \right) } \\ & = \alpha \lim_{n \to \infty} \sqrt[n]{1 + \left( \frac{\beta }{\alpha } \right) ^n}. \end{align*} If $\alpha > \beta $, then $\frac{\beta}{\alpha }<1$ and hence $\left( \frac{\beta }{\alpha } \right) ^n \rightarrow 0$. This implies the result. If $\alpha =\beta $, then \[ \sqrt[n]{\alpha ^n + \beta ^n} = \sqrt[n]{2\alpha ^n} = \alpha \cdot 2^{\frac{1}{n}} \rightarrow \alpha . \] Here, we have used that for any $a>0$, the sequence $\left( a^{\frac{1}{n}} \right) $ converges to $1$. Thus, we proved that \[ \lim_{n \to \infty} \sqrt[n]{\alpha ^n + \beta ^n} = \max\{\alpha ,\beta \} . \]

We can also solve this using the Sandwich's theorem. Again assuming that $\alpha \geq \beta $. Note that \[ \alpha \leq \sqrt[n]{\alpha ^n + \beta ^n} \leq \sqrt[n]{\alpha ^n + \alpha^n}= \alpha \cdot 2^{\frac{1}{n}}. \] Thus, using the Sandwich's theorem the conclusion followed.