3
$\begingroup$

I need to compute:

$\lim \limits_{n \to \infty} \frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot\frac{1}{n^2}$

I tried this

$\lim \limits_{n \to \infty} \frac{n}{n}\frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot\frac{1}{n^2}=\lim \limits_{n \to \infty} \frac{\sqrt[4]{1+\frac{4}{n^3}}\,-\,\sqrt[3]{1+\frac{3}{n^2}}}{\sqrt[5]{1+\frac{1}{n^5}}\,-\, \sqrt[5]{1+\frac{1}{n^4}}}\cdot\frac{1}{n^2}$

I get stuck and I will have to probably use different method, would someone give me an advice, how to approach to this problem?

I also thought about multiplying it by $\frac{(a+b)}{(a+b)}$, but I am not really sure how.

  • 2
    Can you use a Taylor expansion of $\sqrt[k]{1+x}$? – 2017-01-19
  • 0
    Actually I don't know. I am trying to help one of my friend with this example and he didn't tell me. – 2017-01-19
  • 0
    I haven't ever compute the limit with taylor expansion, that it's interesting approach – 2017-01-19
  • 0
    Can you maybe give me some example? – 2017-01-19
  • 1
    Taylor expansion and the generalized binomial theorem are almost the same. Taylor expansion is a very nice way for all these problems with limits. – 2017-01-19
  • 0
    Example is in the answer below. The Taylor expansion gives $\sqrt[k]{1+x} = 1 + \frac{1}{k} x + O(x^2)$ (here the first order expansion suffices). Plugging in the values for $k$ and $x$ in each of the four roots, subtracting in numerator and denominator, then gives the limit. – 2017-01-19
  • 0
    @ClaudeLeibovici in this limit could i change variable like that $t = 1/n$? or i can't because the only limit that admits a succession is $\infty$? – 2017-01-19
  • 1
    @M41Npain. Yes, for sure. This is the way to use. Cheers :-) – 2017-01-19

2 Answers 2

3

The answer is 5.

I expect you're supposed to use the generalized binomial theorem here: namely, $(1+x)^n = \sum_{k=0}^{\infty}\left(\frac{n^{\underline{k}}}{k!}x^k\right)$, where $n^{\underline{k}}$ means $1$ for $k=0$, or $n(n-1)(n-2)\cdots(n-k+1)$ otherwise.

$$\lim \limits_{n \to \infty} \frac{\sqrt[4]{1+\frac{4}{n^3}}\,-\,\sqrt[3]{1+\frac{3}{n^2}}}{\sqrt[5]{1+\frac{1}{n^5}}\,-\, \sqrt[5]{1+\frac{1}{n^4}}}\cdot\frac{1}{n^2} = $$ $$\lim \limits_{n \to \infty} \frac{\left(1 +\frac{1}{n^3} + \mathrm{O}(n^{-6}) \right)\,-\,\left(1+\frac{1}{n^2} + \mathrm{O}(n^{-4})\right)}{\left(1+\frac{1}{5n^5}+\mathrm{O}(n^{-10})\right)\,-\, \left({1+\frac{1}{5n^4} +\mathrm{O}(n^{-8})}\right)}\cdot\frac{1}{n^2} =$$ $$\lim \limits_{n \to \infty} \frac{5 + \mathrm{O}(n^{-3}) \,-\,5n + \mathrm{O}(n^{-1})}{1+\mathrm{O}(n^{-5})\,-\, {n +\mathrm{O}(n^{-3})}} =$$ $$\lim_{n\to\infty}\frac{5n}{n} = 5$$

  • 0
    Okay, I haven't ever used this method. I will check the theorem on the internet. Thank you – 2017-01-19
  • 1
    I'll put the statement in the answer. – 2017-01-19
5

I know you are all "begeistert" with taylor series, but here is another method.

Note that $$\sqrt[3]{n^3+3n}-n=\frac{3n}{(\sqrt[3]{n^3+3n})^2+(\sqrt[3]{n^3+3n})n+n^2}$$ so it follows easily that $$n(\sqrt[3]{n^3+3n}-n)\to \frac{3}{3}=1$$

in a similar fashion one sees that

$$n^2(\sqrt[4]{n^4+4n}-n)\to 1$$ $$n^4(\sqrt[5]{n^5+1}-n)\to \frac{1}{5}$$ $$n^3(\sqrt[5]{n^5+n}-n)\to \frac{1}{5}$$

Now write the original expression as

$$\frac{\frac{n^2(\sqrt[4]{n^4+4n}-n)}{n}-n(\sqrt[3]{n^3+3n}-n)}{\frac{n^4(\sqrt[5]{n^5+1}-n)}{n}-n^3(\sqrt[5]{n^5+n}-n)}\to\frac{0-1}{0-\frac{1}{5}}=5$$

  • 0
    Elementary solutions are the best solutions. +1. – 2017-01-20
  • 0
    @AlfredYerger My sentiments exactly. – 2017-01-20
  • 0
    Never expected this simple a solution +1 – 2017-01-20