If $a,b,c$ are real and nonzero, then $\underset{x\to \infty }{Lt}{\left[\frac{a^{\frac{1}{x}}+b^{\frac{1}{x}}+c^{\frac{1}{x}}}{3}\right]}^x=$

$\underset{x\to \infty }{Lt}{\left[\frac{a^{\frac{1}{x}}+b^{\frac{1}{x}}+c^{\frac{1}{x}}}{3}\right]}^x$

$=e^{\underset{x\to \infty }{Lt}[\frac{a^{\frac{1}{x}}-1}{\frac{1}{x}}+\frac{b^{\frac{1}{x}}-1}{\frac{1}{x}}+\frac{c^{\frac{1}{x}}-1}{\frac{1}{x}}]\times \frac{1}{3}}$

$\therefore \underset{x\to 0}{Lt}\frac{a^x-1}{x}={\log }_{e}a$

$=e^{\underset{\frac{1}{x}\to \infty }{Lt}[\frac{a^{\frac{1}{x}}-1}{\frac{1}{x}}+\frac{b^{\frac{1}{x}-1}}{\frac{1}{x}}+\frac{c^{\frac{1}{x}}-1}{\frac{1}{x}}]\times \frac{1}{3}}$

$=e^{\frac{\log a+\log b+\log c}{3}}$

$\Rightarrow e^{\frac{1}{3}{\log }_{e}abc}={\left(abc\right)}^{\frac{1}{3}}$