1. El_Arrow

2. El_Arrow

i dont understand why she did that

3. anonymous

divide top and bottom by $$n^2$$ so you can more easily see which terms dominate

4. El_Arrow

but if you do that in the bottom for example n^3/n^2 you get n not 1/n

5. anonymous

$$\frac{3n^2+2n}{\sqrt{n^3+n^2+1}}\cdot\frac{1/n^2}{1/n^2}=\frac{3n^2+2n}{\sqrt{n^3+n^2+1}}\cdot\frac{1/n^2}{\sqrt{1/n^4}}=\frac{3+2/n}{\sqrt{1/n+1/n^2+1/n^4}}$$ so you can clearly

6. El_Arrow

oh so you square it when there is a square root?

7. anonymous

Actually he did that..

8. anonymous

Oh sorry, that was someone else you did that..

9. anonymous

see that in the limit as $$n\to\infty$$, the square root tends to $$0$$ while the top tends to $$3$$; this tells us the terms 'blow up' as $$n$$ grows larger and larger since the denominator gets ever smaller

10. anonymous

$\sqrt{\frac{1}{x^2}} = \frac{1}{x} = \sqrt{\frac{1}{x^2}}$

11. anonymous

*who..

12. anonymous

When you take x inside squares, it becomes $$x^2$$..

13. anonymous

*square root brackets..

14. El_Arrow

oh okay i see

15. El_Arrow

i have one more question

16. El_Arrow

where did the 8 in this problem come from?

17. anonymous

the distance between the vertex and focus ($$-4$$) is equal to the distance between the vertex and directrix

18. El_Arrow