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This question already has an answer here:

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

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

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

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marked as duplicate by Arjang, JonMark Perry, Guy Fsone, Fabio Lucchini, muaddib Jan 28 '18 at 12:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

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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.

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

details : Limit Computation, Sandwich.

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  • $\begingroup$ I think it is better to add few steps to show how you obtain this answer. Otherwise it is not helpful. $\endgroup$ – Bumblebee Apr 2 '17 at 22:09
  • $\begingroup$ @Nil : linked to details, couldn't find the equivalent one for min. $\endgroup$ – Arjang Apr 2 '17 at 22:26

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