Question

Real Analysis...
Prove directly from the definition that
lim n^2/n+1 =1

Answer 1

limit as n goes to what?

Answer 2

If there's no error in that question, it would be a really weird limit...

Answer 3

Is it
\[ \frac{n^2}{n} + 1 \]
or
\[\frac{n^2}{n+1} \]
?

Answer 4

i bet it is
\[\lim_{n\to \infty}\frac{n^2}{n^2+1}\]

Answer 5

\[\lim \frac{ n^{2} }{ n^{2}+1 }=1\]

This is similar to something we did in class so I modeled it after that but I don't know if its right. This is what I have:

PROOF:

Let eplison >0.

\[\left| \frac{ n^{2} }{ n^{2}+1 }- 1 \right|= \left| \frac{ -n-1 }{ n^{2}+1 } \right|= \frac{ n+1 }{ n^{2}+1 }\]

Replace n+1 with something larger; 2n

\[\frac{ n+1 }{ n^{2}+1 }<\frac{ 2n }{ n^{2}+1 }\]

then replace n^2+1 by something smaller; n^2

\[\frac{ 2n }{ n^2 +1} <\frac{ 2n }{ n^2 }=\frac{ 2 }{ n }\]

Thus,
\[\left| \frac{ n^2 }{ n^2 +1}-1 \right|<\frac{ 2 }{ n }\]

Now, choose N, so N>2/epsilon.
Then, if \[n \ge N\],

\[\left| \frac{ n^2 }{ n^2 +1 }-1 \right| <\frac{ 2 }{ n }<\frac{ 2 }{ N }< \epsilon \]

Answer 6

I'm horrible with proofs and I'm really not confident that this is right....

Answer 7

what is the original question?

Answer 8

\[\frac{n^2}{n^2+1}-1=\frac{n^2-(n^2+1)}{n^2+1}=\frac{-1}{n^2+1}\]

Answer 9

2a

Answer 10

Well, for one
\[ \left|\frac{n^2}{n^2+1}-1\right| \neq \frac{n+1}{n^2+1} \]
but
\[ \left|\frac{n^2}{n^2+1}-1\right| = \frac{1}{n^2+1} \]

Answer 11

Your proof looks good in principle but the algebra has an error, as the gentleman above has pointed out.

Answer 12

so let \(\epsilon>0\) and you want \(\frac{1}{n^2+1}<\epsilon\) so take \(n^2+1>\frac{1}{\epsilon}\) which you can do by whatever that thing is called that says the natural numbers are not bounded above

Answer 13

or if you want to be annoyingly pedantic, take
\[n>\sqrt{\frac{1}{\epsilon}-1}\]

Answer 14

Ok... I think I get this now. Im going to go work on it.
Thank you all for your help!!!