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I'm trying to solve:

$\hspace{40mm}$ $\lim_{n\to\infty}$ $(\frac{\sqrt{n^4-2n^3} - n^2}{10n+2})^2$

This is what I did to try to solve it, but my answer doesn't match the possible answers so I know it's wrong.

$\lim_{n\to\infty}$ $(\frac{\sqrt{n^4-2n^3} - n^2}{10n+2})^2$

  • Highest power of denominator is n, so pull out $n^2$ from the square root to give an n

$\lim_{n\to\infty}$ $(\frac{n(\sqrt{n^2-2n} )- n^2}{10n+2})^2$

  • Factor out n from numerator and denominator

$\lim_{n\to\infty}$ $(\frac{n((\sqrt{n^2-2n})- n)}{n(10+\frac{2}{n})})^2$

  • Cancel n from numerator and denominator

$\lim_{n\to\infty}$ $(\frac{\sqrt{n^2-2n} - n}{10+\frac{2}{n}})^2$

  • Take $n^2$ out from the square root

$\lim_{n\to\infty}$ $(\frac{n\sqrt{1-\frac{2}{n}} - n}{10+\frac{2}{n}})^2$

  • Factor out n from numerator

$\lim_{n\to\infty}$ $(\frac{n((\sqrt{1-\frac{2}{n}}) - 1)}{10+\frac{2}{n}})^2$

At this point I would take the limit to be 0, but the possible answers are 0.05, 0.01, 0.001, or 0.1. Where did I go wrong?

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  • $\begingroup$ Do you want $n\to\infty$ instead? $\endgroup$ – user9464 Oct 11 '19 at 16:41
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You have got everything right till the end. Use the fact that $(1-\frac 2 n)^{1/2} =1-\frac 1 n +o(\frac 1 n)$ to see that the limit is $(0.1)^{2}=0.01$.

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  • $\begingroup$ Wouldn't I still have the n at the front of the numerator which would go to infinity? Also, I'm not sure what the o($\frac{1}{n}$) is and how the two are equal. $\endgroup$ – Edgar Smith Oct 12 '19 at 19:59
  • $\begingroup$ $a_n=o(b_n)$ means $\frac {a_n} {b_n} \to 0$. You get $(1-\frac 2 n)^{1/2}=1-\frac 1n +o(\frac 1 n)$ from the expansion of $(1-x)^{1/2}$. @EdgarSmith $\endgroup$ – Kavi Rama Murthy Oct 12 '19 at 23:16
  • $\begingroup$ Ah ok, thanks. But wouldn't there still be the n at the front of the numerator, which would go to infinity? $\endgroup$ – Edgar Smith Oct 13 '19 at 16:50
  • $\begingroup$ @EdgarSmith $n(1-\frac 1 n +o(\frac 1 n)-1)=-1+o(1) \to 1$. $\endgroup$ – Kavi Rama Murthy Oct 13 '19 at 23:12
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set $1/n=h$

$$\sqrt{n^4-2n^3}=\sqrt{\dfrac{1-2h}{h^4}}=\dfrac{\sqrt{1-2h}}{h^2}$$

Rationalize the numerator

$$\lim_{h\to0^+}\dfrac{\sqrt{1-2h}-1}{h(10+2h)}=\lim_{h\to0^+}\dfrac{{1-2h}-1}{h(10+2h)(\sqrt{1-2h}+1)}=-\dfrac2{10(1+1)}$$

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