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How do i prove the limit below? I’ve tried, but i got nothing.

$ \lim\limits_{n\to \infty} (3^n + 4^n)^{1/n} = 4. $

Thanks for any help.

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In general, let $\alpha_1,\alpha_2,\dots,\alpha_m$ be positive numbers. Let $A=\max\limits_{1\leq i\leq m}{\alpha_i}$

Then $$A^n\leq \alpha_1^n+\cdots+\alpha_m^n\leq mA^n$$

Thus $$A\leq (\alpha_1^n+\cdots+\alpha_m^n)^{1/n}\leq m^{1/n} A$$

So $$\lim_{n\to\infty} (\alpha_1^n+\cdots+\alpha_m^n)^{1/n}=\max\limits_{1\leq i\leq m}{\alpha_i}$$

since $m^{1/n}\to 1$.

Hint 2: $4^n < 3^n + 4^n < 2\cdot 4^n$

How about:

$$=\lim_{n \to \infty} 4\left(1+\frac{3^n}{4^n}\right)^{1/n} = \lim_{n \to \infty} 4\left(1+0.75^n\right)^{1/n} = 4$$

Hint: as $n \to \infty$, $4^n$ becomes much larger than $3^n$.

$$(3^n+4^n)^{1/n}= e^\frac{\ln(3^n+4^n)}{n}= e^\frac{\ln(4^n(1+(3/4)^n)}{n}= e^{\ln(4)}e^{\frac{1}{n}\ln(1+(3/4)^n)}\longrightarrow_{n\to \infty} 4 $$

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