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Mathematics 11 Online
OpenStudy (el_arrow):

don't understand this limit problem please help

OpenStudy (el_arrow):

OpenStudy (el_arrow):

i dont understand why she did that

OpenStudy (anonymous):

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

OpenStudy (el_arrow):

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

OpenStudy (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

OpenStudy (el_arrow):

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

OpenStudy (anonymous):

Actually he did that..

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

\[\sqrt{\frac{1}{x^2}} = \frac{1}{x} = \sqrt{\frac{1}{x^2}}\]

OpenStudy (anonymous):

*who..

OpenStudy (anonymous):

When you take x inside squares, it becomes \(x^2\)..

OpenStudy (anonymous):

*square root brackets..

OpenStudy (el_arrow):

oh okay i see

OpenStudy (el_arrow):

i have one more question

OpenStudy (el_arrow):

where did the 8 in this problem come from?

OpenStudy (anonymous):

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

OpenStudy (el_arrow):

so you are add -4+-4?

OpenStudy (anonymous):

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