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?