Evaluating $ \lim_{n \to \infty} (2^n + 6^n)^{1/n}$

Question

I'm completely stuck on how I would go about solving this problem. I'm not sure if I need to use e or ln. Could someone give hints on how I would go about solving this problem?

$$ \lim_{n \to \infty} (2^n + 6^n)^{1/n}$$

Answer 1

We may transform the above expression and utilize $e=\lim_{n\rightarrow\infty}(1+\dfrac{1}{n})^n$.

$$\lim_{n\rightarrow\infty}(2^n+6^n)^\frac{1}{n}$$ $$=\lim_{n\rightarrow\infty}(6^n)^\frac{1}{n}(1+\frac{1}{3^n})^\frac{1}{n}$$ $$=6\lim_{n,3^n\rightarrow\infty}[(1+\frac{1}{3^n})^{3^n}]^\frac{n}{3^n}$$ $$=6\lim_{3^n\rightarrow\infty}[(1+\frac{1}{3^n})^{3^n}]^{\lim_{n,3^n\rightarrow\infty}\frac{n}{3^n}}$$ $$=6e^0=6$$

Answer 2

Consider $\lim_{n \to \infty} (a^n + b^n)^{1/n} $ where $0 < a < b$.

Let $f(n) =(a^n + b^n)^{1/n} $.

Then $f(n) \gt(b^n)^{1/n} =b$.

Also

$\begin{array}\\ f(n) &=(a^n + b^n)^{1/n}\\ &=b(1+(a/b)^n)^{1/n}\\ &=b(1+r^n)^{1/n} \qquad r=a/b < 1\\ \end{array} $

If $(1+r^n)^{1/n} =1+h$ then $1+r^n =(1+h)^n \ge 1+hn $ by Bernoulli's inequality so $r^n \gt hn $ or $h \lt \dfrac{r^n}{n} $.

Rearranging, $(1+r^n/n)^n \ge 1+r^n $ so $1+r^n/n \ge (1+r^n)^{1/n} $.

Therefore

$\begin{array}\\ f(n) &=b(1+r^n)^{1/n}\\ &\le b(1+r^n/n)\\ &= b+br^n\\ \end{array} $

so $b \le (a^n + b^n)^{1/n} \le b+b(a/b)^n/n $.

Since $\lim_{n \to \infty} b(a/b)^n/n =0$, $\lim_{n \to \infty} (a^n + b^n)^{1/n} =b$.