Crafty Infinite Limit

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L & = \lim_{x \to \infty} \left(\frac {2^\frac 1x + 27^\frac 1x + 8^\frac 1x}3\right)^x & \small \color{#3D99F6} \text{Let }u = \frac 1x \\ & = \lim_{u \to 0} \left(\frac {2^u + 27^u + 8^u}3\right)^\frac 1u & \small \color{#3D99F6} \text{As }\lim_{x\to a} f(x)^{g(x)} = 1^\infty \\ & = \exp \left(\lim_{u \to o} \frac 1u \left(\frac {2^u + 27^u + 8^u}3- 1\right)\right) & \small \color{#3D99F6} \implies \lim_{x\to a} f(x)^{g(x)} = e^{\lim_{x\to a} g(x)(f(x)-1)} \\ & = \exp \left(\lim_{u \to o} \frac {2^u + 27^u + 8^u-3}{3u} \right) & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \exp \left(\lim_{u \to o} \frac {2^u\ln 2 + 27^u\ln 27 + 8^u\ln 8}3 \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t }u. \\ & = \exp \left(\frac {\ln 2}3 + \ln 3 + \ln 2\right) \\ & = \boxed{6\sqrt[3]2} \end{aligned}$