How to solve this particular indetermination: $0*\infty$

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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 $4.99999993750000000$.