Question

Prove directly from the definition that:
lim n^2/(n^(2)+1) = 1

Answer 1

you need to prove

\[\frac{\lim_{n \rightarrow ?} n^2}{\lim_{n \rightarrow ?} n^2 + 1}\]

Answer 2

I am not good with proofs, so I am having a really hard time with this. Could you guide me?

Answer 3

so you need to look at

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

which is tha same as

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

which simplifies to

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

which will allow you to find the limit...

Answer 4

Ok...so where do I go from here?

Answer 5

I see how you got that. We did a problem similar to this in class, so that part clicked,but we never finished it. So this is where I am stuck even more lol

Answer 6

What does the limit go to?

Answer 7

it doesn't specify

Answer 8

i think infinity though

Answer 9

it must be infinity, otherwise this is never 1

Answer 10

thought so

Answer 11

Okay, so use the definition of limit to solve

Answer 12

you can also divide top and bottom by \(n^2\) to get the same result, although @campbell_st method is more elegant

Answer 13

ok, so i am suppose to show my work for the definition of limit to solve but idk how to do that, because I just know how to look at it and just KNOW, i dont know how i am suppose to prove

Answer 14

this is the limit of a squence. right?
\[ \large \lim_{n\to\infty}\frac{n^2}{n^2+1}=1 \]

Answer 15

yes

Answer 16

let's remember the definition. left \(a_n=n^2/(n^2+1)\) then
\[ \large \lim a_n=1 \]
if and only if
\[ \large (\forall\varepsilon>0)(\exists N)(n\geq N\to|a_n-1|<\varepsilon) \]
ok?

Answer 17

let \(\varepsilon>0\) then
\[ \large |a_n-1|=\left|\frac{n^2}{n^2+1}-1\right|=
\left|\frac{-1}{n^2+1}\right|=\frac{1}{n^2+1}<
\varepsilon \]
then
\[ \large \frac{1}{\varepsilon}<n^2+1 \]
\[ \large \frac{1}{\varepsilon}-1<n^2 \]
\[ \large n>\sqrt{\frac{1}{\varepsilon}-1} \]

Answer 18

so let
\[ \large N=\sqrt{\frac{1}{\varepsilon}-1} \]