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The limit in question is: $$\lim_{\color{red}n\to\infty} 2n\left(\sqrt{n^6+5n^2}-n^3\right)$$

By looking it up on wolfram alpha I found out the answer is 5 but I am not so sure how to arrive to it. I tried to change the indetermination to infinity/infinity and apply L'Hopital's rule but to no avail. Thanks in advance.

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$\begin{align}\lim_{n\to\infty}2n(\sqrt{n^6+5n^2}-n^3)&=\lim_{n\to\infty}2n(\sqrt{n^6+5n^2}-n^3)\cdot\frac{(\sqrt{n^6+5n^2}+n^3)}{\sqrt{n^6+5n^2}+n^3}\\&=\lim_{n\to\infty} \frac{2n(n^6+5n^2-n^6)}{\sqrt{n^6+5n^2}+n^3}\\&=\lim_{n\to\infty}\frac{10n^3}{\sqrt{n^6+5n^2}+n^3}\\&=\lim_{n\to\infty}\frac{10n^3}{n^3(\sqrt{1+\frac{5}{n^4}}+1)}\\&=5\end{align}$

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  • $\begingroup$ Thanks for the answer. This is the easiest way to solve it. $\endgroup$ – Leo Lerena May 27 '15 at 16:46
  • $\begingroup$ Glad to help you! $\endgroup$ – Lucas May 27 '15 at 17:42
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For the limit itself, André Nicolas gave the good hint.

You can do a bit more using Taylor $$A=2n\left(\sqrt{n^6+5n^2}-n^3\right)=2 n^4\left(\sqrt{1+\frac5{n^4}}-1\right)$$ Now, us Taylor for $\sqrt{1+x}=1+\frac x2-\frac {x^2}8 +\cdots$ and replace $x$ by $\frac 5{n^4}$. You then arrive to $$A=2n^4\big(1+\frac{5}{2 n^4}-\frac{25}{8 n^8}+\cdots-1\Big)=5-\frac{25}{4 n^4}+\cdots$$ which shows the limit and how it is approached.

For illustration purposes, try with $n=10$; the original expression evaluates $\approx 4.999375156$ while the approximation formula gives $4.999375000$. For $n=100$, the corresponding results would be $\approx 4.99999993750000156$ and $4.99999993750000000$.

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  • $\begingroup$ in my opinion you have to have the best answer! anyway nice method $\endgroup$ – Lucas May 27 '15 at 6:05
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    $\begingroup$ Thank you very much ! I started my love for Taylor series almost sixty years ago. What I enjoy is that you can not only solve easily the problem of the limit but also find the asymptotic behavior and this is extremely useful in applied mathematics. $\endgroup$ – Claude Leibovici May 27 '15 at 6:12
  • $\begingroup$ Sorry. This is the best answer in terms of clarity. But I don't really master Taylor so it didn't help me particulary. Thanks anyways for the effort. $\endgroup$ – Leo Lerena May 27 '15 at 16:42
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    $\begingroup$ Don't worry ! You will learn soon and remember my words : using Taylor, you can do almost everything and even more ! Cheers :-) $\endgroup$ – Claude Leibovici May 27 '15 at 19:18
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Take n^6 out of square root and then n^3 from bracketed terms. U will be left with n^4 in numerator.

Now multiply and divide with conjugate of bracketed term left in above operation. U will get 10/2 = 5.

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  • $\begingroup$ It's not easily understood what you wrote because of the syntax but I found your answer the most straightfoward. $\endgroup$ – Leo Lerena May 27 '15 at 5:35
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    $\begingroup$ I'm new to stack exchange, lea $\endgroup$ – Amit K Anand May 27 '15 at 5:36
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    $\begingroup$ I find it difficult to use mathematical symbols $\endgroup$ – Amit K Anand May 27 '15 at 5:37
  • $\begingroup$ If you wrap your math functions in single $ (for content you want in line) and double dollar signs are for new line content, it will improve your readability. More complex equations use Latex and there are various examples available on the web. You can also click the edit button on any answer on this page and see how the others format their content as well (just hit cancel when you are done looking). $\endgroup$ – scrappedcola Jun 26 '15 at 18:40

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