Evaluate $\lim_{n \to \infty} \sqrt[n]{3^n+4^n}$

Question

$$\lim_{n \to \infty} \sqrt[n]{3^n+4^n}$$

Is there there a way to solve this without using $e^{ln(3^n+4^n)}$?

Maybe: $\displaystyle\lim_{n \to \infty} \sqrt[n]{4^n}=4\,\leq\,\lim_{n \to \infty} \sqrt[n]{3^n+4^n}\,\leq\,\lim_{n \to \infty} \sqrt[n]{2\cdot4^n}=4$?

Links to some generalizations: How to show $\lim_{n \to \infty} \sqrt[n]{a^n+b^n}=\max {a,b}$? (and other posts linked there) or Convergence of $\sqrt[n]{x^n+y^n}$ (for $x, y > 0$) (and other posts linked there.) — Martin Sleziak, Jan 28 '18 at 09:13

score 5 · Accepted Answer · answered Apr 02 '17 at 21:03

5

$$\lim _{ n\rightarrow \infty }{ \sqrt [ n ]{ 3^{ n }+4^{ n } } } =4\cdot \lim _{ n\rightarrow \infty }{ \sqrt [ n ]{ \left( \frac { 3 }{ 4 } \right) ^{ n }+1 } } =4$$

answered Apr 02 '17 at 21:03

haqnatural

22,026

score 2 · Answer 2 · answered Apr 02 '17 at 21:23

Your proof is fine, provided you can use that $$ \lim_{n\to\infty}\sqrt[n]{2}=1 $$ This follows from Bernoulli’s inequality $(1+x)^n\ge 1+nx$, whenever $x>-1$ and $n$ is a positive integer, in the form $$ \sqrt[n]{1+nx}\le 1+x $$ For $x=1/n$ this reads $$ \sqrt[n]{2}\le 1+\frac{1}{n} $$ and therefore, from $$ 1\le\sqrt[n]{2}\le 1+\frac{1}{n} $$ and the squeeze theorem, you can conclude.

Then your application of the squeeze theorem to $$ 4=\sqrt[n]{4^n}\le\sqrt[n]{3^n+4^n}\le \sqrt[n]{2\cdot 4^n}=4\sqrt[n]{2} $$ is good.

score 1 · Answer 3 · edited Apr 13 '17 at 12:21

1

$\lim_{n \to \infty} \sqrt[n]{a^n+b^n}= \text{max}(a,b)$

details : Limit Computation, Sandwich.

edited Apr 13 '17 at 12:21

Community

1

answered Apr 02 '17 at 21:49

jimjim

9,855

I think it is better to add few steps to show how you obtain this answer. Otherwise it is not helpful. – Bumblebee Apr 02 '17 at 22:09
@Nil : linked to details, couldn't find the equivalent one for min. – jimjim Apr 02 '17 at 22:26

Evaluate $\lim_{n \to \infty} \sqrt[n]{3^n+4^n}$

3 Answers3

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