We assume that $a\ge 0$ and $b\ge 0$ and write
$$\begin{align}
\lim_{x \to \infty} \left(\frac{a^x+b^x}{2}\right)^{1/x} &=\,a\lim_{x \to \infty} \left(\frac{1+(b/a)^x}{2}\right)^{1/x}
\end{align}$$
If $a=b$, then the limit is obviously equal to $a$.
Now, we assume without loss of generality that $a>b$. Then, it is easy to see that
$$\begin{align}
\lim_{x \to \infty} \left(\frac{a^x+b^x}{2}\right)^{1/x} &=\,a\lim_{x \to \infty} \left(\frac{1+(b/a)^x}{2}\right)^{1/x}\\\\
&=a\lim_{x \to \infty} \exp\left(\log \left(\frac{1+(b/a)^x}{2}\right)^{1/x} \right)\\\\
&=a \exp\left(\lim_{x \to \infty} \frac1x \,\log \left(\frac{1+(b/a)^x}{2}\right) \right)\\\\
&=a \exp\left(\lim_{x \to \infty} \left(\frac1x\right) \times \lim_{x \to \infty} \log \left(\frac{1+(b/a)^x}{2}\right) \right)\\\\
&=a \exp\left(0 \times \log \left(\frac12\right) \right)\\\\
&=a
\end{align}$$
Of course, one might have observed that inasmuch as (1) $b/a<1$, and (2) $y^{1/x} \to 1$ as $x \to \infty$ for $0<y<1$, then (3) $(b/a)^{x}$ approaches zero as $x \to \infty$ and thus (4) $(\frac{1+(b/a)^x}{2})^{1/x} \to 1$ as $x \to \infty$.
Note that if the limit were $x \to -\infty$, then we proceed analogously.
We assume that $a\ge 0$ and $b\ge 0$ and write
$$\begin{align}
\lim_{x \to -\infty} \left(\frac{a^x+b^x}{2}\right)^{1/x} &=\,b\lim_{x \to -\infty} \left(\frac{1+(a/b)^x}{2}\right)^{1/x}
\end{align}$$
If $a=b$, then the limit is obviously equal to $b$.
Now, we assume without loss of generality that $a>b$. Then, it is easy to see that as $x \to -\infty$, the term $(a/b)^x$ goes to zero and $(1/2)^{1/x} \to 1$. Thus,
$$\lim_{x \to -\infty} \left(\frac{a^x+b^x}{2}\right)^{1/x} =b$$