Find the lim

Relevant wiki: L'Hôpital's Rule

$\begin{aligned} L & = \lim_{x \to 0} \left(\frac {2^x+3^x+5^x}3\right)^\frac 3x & \small \color{#3D99F6} \text{A }1^\infty \text{ case} \implies \\ & = \exp \left(\lim_{x \to 0} \frac 3x \left(\frac {2^x+3^x+5^x}3 - 1\right)\right) & \small \color{#3D99F6} \lim_{x \to a}(f(x))^{g(x)} = e^{\lim_{x \to a}g(x)(f(x)-1)} \\ & = \exp \left(\lim_{x \to 0} \frac {2^x+3^x+5^x-3}x \right) & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \exp \left(\lim_{x \to 0} \frac {2^x\ln 2+3^x\ln 3+5^x\ln 5}1 \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x. \\ & = \exp \left(\ln 2+\ln 3+\ln 5\right) \\ & = e^{\ln 30} = \boxed{30} \end{aligned}$

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Reference: Limit of $1^\infty$ -case (see 2nd method )

$\begin{aligned} L &= \lim_{x \to 0} \left( \dfrac{2^x + 3^x + 5^x}{3} \right)^{\dfrac{3}{x}} \\ \\ \\ \ln L &= \ln \left( \lim_{x \to \infty} \left( \dfrac{2^x + 3^x + 5^x}{3} \right)^{\dfrac{3}{x}} \right) \\ \\ \\ &= \lim_{x \to 0} \left( \ln \left( \dfrac{2^x + 3^x + 5^x}{3} \right)^{\dfrac{3}{x}} \right) \\ \\ \\ &= \lim_{x \to 0} \left( \dfrac{3}{x} \ \ln \left( \dfrac{2^x + 3^x + 5^x}{3} \right) \right) \\ \\ \\ &= \lim_{x \to 0} \dfrac{3 \ln \left( \dfrac{2^x + 3^x + 5^x}{3} \right)}{x} \qquad \qquad \small \text{[Indeterminate form } \frac{0}{0} \text{; use L'Hospital's Rule]}\\ \\ \\ &= \lim_{x \to 0} \dfrac{3 \cdot \dfrac{1}{3} \left( 2^x \ln 2 + 3^x \ln 3 + 5^x \ln 5 \right)}{\left( \dfrac{2^x + 3^x + 5^x}{3} \right)} \\ \\ \\ &= \ln 2 + \ln 3 + \ln 5 = \ln 30 \\ \\ \\ \therefore L &= \boxed{30} \end{aligned}$