Prove the following limit $ \lim_{n\to \infty} (3^n + 4^n)^{1/n} = 4 $
Solution 1:
Hint 2: $4^n < 3^n + 4^n < 2\cdot 4^n$
Solution 2:
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 $m^{1/n}\to 1$.
Solution 3:
How about: $$=\lim_{n \to \infty} 4\left(1+\frac{3^n}{4^n}\right)^{1/n} = \lim_{n \to \infty} 4\left(1+0.75^n\right)^{1/n} = 4$$