# Finding a limit using AM-GM?

Doing some calculus papers before going university and I found this question:

Find $$\lim_{x\to\infty} \left[\frac{1}{3} \left(3^\frac{1}{x} + 8^\frac{1}{x} + 9^\frac{1}{x} \right)\right]^x$$

My solution was to use AM-GM, as shown: Since $$3^\frac{1}{x} + 8^\frac{1}{x} + 9^\frac{1}{x} \geq 3 \sqrt[3]{3^\frac{1}{x} 8^\frac{1}{x} 9^\frac{1}{x}} = 3 \cdot 6^\frac{1}{x}$$, it follows that $$\lim_{x\to\infty} \left[\frac{1}{3} \left(3^\frac{1}{x} + 8^\frac{1}{x} + 9^\frac{1}{x} \right)\right]^x \geq \lim_{x\to\infty} \left[\frac{1}{3}\cdot 3 \cdot 6^\frac{1}{x} \right]^x=6$$. Equality holds as $$\lim_{x \to\infty} 3^\frac{1}{x} = \lim_{x \to\infty} 8^\frac{1}{x} = \lim_{x \to\infty} 9^\frac{1}{x}$$

Is this solution rigorous enough? Or what would be a better way (if any) of solving this?

Thanks

• Your statement 'Equality holds as $\lim_{x \to\infty} 3^\frac{1}{x} = \lim_{x \to\infty} 8^\frac{1}{x} = \lim_{x \to\infty} 9^\frac{1}{x}$' does not make sense. May 27, 2022 at 9:03
• @KaviRamaMurthy my apologies but could I understand why not? all of those 3 limits tend to 1 at infinity May 27, 2022 at 9:17
• I think what OP wants to say is that AM=GM since the limits of all those terms considered in the inequality are equal. May 27, 2022 at 9:19
• Alternatively write the limit from $(f(x))^x$ as $e^{x\ln(f(x))}$ then sub $y=\frac{1}{x}$ and apply L'Hôpital's rule May 27, 2022 at 9:23
• @Doobius But this cannot be used here (at least not without further justification): For any given $x$, we only know the inequality, and raising the inequality to the power of $x$ might destroy equality for $x \to \infty$. May 27, 2022 at 9:24