Prove that this sequence is convergent by using the limit definition Suppose I have the following sequence:
$ a_n = \frac{2n -\sqrt (n^2+1)}{ n+cos(n)}$
The definition says that for each $\epsilon > 0 $ there exists a positive integer $n_0 $ such that for all n > $n_0 : |a_n - L | < \epsilon$
Before starting to prove this I try to find a value for $ n_0 $ in function of $ \epsilon $.
We can write that:
$ a_n = \frac{2n - \sqrt (n^2+1)}{ n+cos(n)} \leq \frac{2n}{ n+cos(n)} $
But then I am stuck with the $ cos(n) $ in the denominator.
My idea was to use the fact that the cosine is always between -1 and 1.
Thus, $ cos(n) \leq 1 $ and $ a_n = \frac{2n - \sqrt (n^2+1)}{ n+cos(n)} \leq \frac{2n}{ n+cos(n)} \leq \frac{2n}{ n+1} \leq \frac{2n}{ n} \leq 2 $
But then again I am stuck because this doesn't help me to find an $ n_0 $ in function of $ \epsilon $
EDIT:
@TheSilverDoe reminded me that I forgot the '-L' part in the definition.
I have to calculate the limit first:
Write:
$ \frac{2n - \sqrt (n^2+1)}{ n +1} \leq \frac{2n - \sqrt (n^2)}{ n+1} \leq \frac{2n - \sqrt (n^2+1)}{ n+cos(n)} \leq\frac{2n - \sqrt (n^2)}{ n-1} \leq\frac{2n - \sqrt (n^2+1)}{ n-1} $
or
$ \frac{n}{ n+1} \leq \frac{2n - \sqrt (n^2+1)}{ n+cos(n)}  \leq\frac{n}{ n-1} $
From that we can easily see that the limit is equal to 1.
Back to my original question we have:
$ a_n - L = \frac{2n - \sqrt (n^2+1)}{ n+cos(n)} - 1 \leq \frac{2n}{ n+cos(n)} - 1 \leq \frac{2n}{ n+1} - 1 \leq \frac{2n}{ n} - 1 \leq 2 - 1 = 1$
This brings me back to the same question I had originally
 A: It doesn't help you to show that $a_n \le 2$ (or $\le 3$ or $\le 35436$) unless you can get close to the $2$.  And as $\sqrt{n^2 + 1} > n$ then $2n  -\sqrt{n^2 + 1} < 2n - n =n$ you cant really get close to $2$.
But we could do $\frac {n-1}{n+1}=\frac {2n-(n+1)}{n+1} = \frac{2n-\sqrt{n^2 + 2n+1}}{n+1}<\frac {2n-\sqrt{n^2+1}}{n+\cos n}= a_n < \frac{2n-\sqrt{n^2}}{n-1} = \frac {n}{n-1}$.
We should just be able to look and $\frac {n-1}{n+1} \to 1$ and $\frac {n}{n-1}\to 1$ and use the squeeze theorem but... where's the fun in that?  Which is my tongue in check way of saying we'll never learn how to not fear delta epsilon proofs if we keep avoid them.
$\frac {n-1}{n+1} = 1-\frac 2{n+1} < 1 < 1 + \frac1{n-1} = \frac n{n-1}$.
So we have both $a_n$ and $1$ between the extremes of $\frac {n-1}{n+1}$ and $\frac n{n-1}$.
So $|1-a_n| < |\frac n{n-1} - \frac {n-1}{n+1}| = \frac 1{n-1} + \frac 2{n+1}< \frac 1{n-1} +\frac 2{n-1} = \frac 3{n-1}$.
So if we want $\frac 3{n-1} < \epsilon$ it is sufficient (more than sufficient) to have $\frac 3\epsilon < n-1$ so let $n_0 = \frac 3\epsilon +1$.
