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This question already has an answer here:

How to calculate $$\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt[n]{{{3^n} + {4^n}}}} \right)?$$


I saw on the suggestions that you can use the sandwich theorem or compression. And the fact that $\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt[n]{2}} \right) = 1$

So I must first find upper bounds and lower.

someone help me with this please?

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marked as duplicate by Guy Fsone, The Phenotype, Namaste calculus Jan 30 '18 at 0:27

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    $\begingroup$ For an upper bound of $4$, notice that $3^n \le 4^n$, so $3^n+4^n \le 2 * 4^n$, and thus $(3^n + 4^n)^\frac{1}{n} \le 4 * 2^{\frac{1}{n}}$. $\endgroup$ – Hetebrij Sep 8 '15 at 9:04
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    $\begingroup$ For the lower boud, notice that $3^n+4^n \ge 4^n$, so $(3^n + 4^n)^\frac{1}{n} \ge 4$. $\endgroup$ – Hetebrij Sep 8 '15 at 9:06
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Complete Solution: Sandwich theorem

$$ \sqrt[n]{4^n} \le \sqrt[n]{4^n + 3^n} \le \sqrt[n]{2\times4^n}$$

and so

$$ 4 \le \sqrt[n]{4^n + 3^n} \le 2^{1/n} \times 4 $$

Since $2^{1/n} \to 1$ as $n \to \infty$, Hence the required limit is $4$.

OR, use a sledgehammer:

Use the fact that for positive $a_n$ if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists then so does $\lim\limits_{n\rightarrow\infty}\root n\of {a_n}$ and they are equal.Here $a_n=4^n+3^n$ and one can show $$ \lim\limits_{n\rightarrow\infty} {4^{n+1}+3^{n+1}\over 4^n+3^n} =4. $$ So, then $\lim\limits_{n\rightarrow\infty} \root n\of{4^n+3^n}=4$ as well.

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Hint.

$$\sqrt[n]{{{3^n} + {4^n}}}=\sqrt[n]{4^n(1+\left(\frac{3}{4}\right)^n)}=4\sqrt[n]{1+\left(\frac{3}{4}\right)^n}$$

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$$3^n + 4^n \sim 4^n \implies \sqrt[n] {3^n + 4^n} \sim \sqrt[n]{4^n} = 4$$

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    $\begingroup$ This magical infinitesimal equivalences seem profoundly antipedagogical to me, e.g. $\lim_{x\to\infty}[ (3^n+4^n) -4^n]$ ~ $\lim_{x\to\infty} (4^n -4^n )=0$ $\endgroup$ – Miguel Sep 8 '15 at 9:22
  • $\begingroup$ @MiguelAtencia Well you've got to lear to use them, and understand why sometimes they fail, but they are very powerful once you do. I also find them useful to get a better grasp of limits; to not only treat them with a rigid set of rules, but to understand (intuitively) that $4^n$ is so much bigger than $3^n$ that you can (sometimes) forget the latter one. $\endgroup$ – Ant Sep 8 '15 at 10:29
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You can also do the following:

$$ (3^n+4^n)^{\frac{1}{n}}=e^{\frac{1}{n}\ln(3^n+4^n)}=e^{\frac{1}{n}\ln[4^n(1+\frac{3^n}{4^n})]}$$ $$ = e^{\frac{1}{n}[n\ln 4 + \ln (1+\frac{3^n}{4^n})]}=e^{\ln 4 + \frac{\ln (1+\frac{3^n}{4^n})}{n}}=4\cdot e^{\frac{\ln (1+\frac{3^n}{4^n})}{n}}$$

It is now easy to show that $e^{\frac{\ln (1+\frac{3^n}{4^n})}{n}}$ goes to $0$ for $ n \rightarrow \infty $.

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