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

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?

• Do you want $n\to\infty$ instead? – user9464 Oct 11 '19 at 16:41

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 $$(0.1)^{2}=0.01$$.

• 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. – Edgar Smith Oct 12 '19 at 19:59
• $a_n=o(b_n)$ 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 – Kavi Rama Murthy Oct 12 '19 at 23:16
• Ah ok, thanks. But wouldn't there still be the n at the front of the numerator, which would go to infinity? – Edgar Smith Oct 13 '19 at 16:50
• @EdgarSmith $n(1-\frac 1 n +o(\frac 1 n)-1)=-1+o(1) \to 1$. – Kavi Rama Murthy Oct 13 '19 at 23:12

set $$1/n=h$$

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

Rationalize the numerator

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