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

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?

marked as duplicate by Guy Fsone, The Phenotype, Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 30 '18 at 0:27

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

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.

Hint.

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

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

• This magical infinitesimal equivalences seem profoundly antipedagogical to me, e.g. $\lim_{x\to\infty}[ (3^n+4^n) -4^n]$ ~ $\lim_{x\to\infty} (4^n -4^n )=0$ – Miguel Sep 8 '15 at 9:22
• @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. – Ant Sep 8 '15 at 10:29

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 $n \rightarrow \infty$.