Prove the limit.. $\varinjlim \sqrt{n^2+1}-n=0$.
I need to prove that this converges to 0. 
Usung the definition of a sequence helps for the normal problems but for this I believe the triangle inequality is used at some point....
I let $S_n=\varinjlim \sqrt{n^2+1}-n$
Then $|S_n -S|<\epsilon \rightarrow |\varinjlim \sqrt{n^2+1}-n-0|<\epsilon$.
I get to $n^2+1<(\epsilon+n)(\epsilon+n)$ and I'm stuck..
By the definition I'm trying to find $n>N$....
How can I solve this by contradiction? ( since the above lead me nowhere)

Try this:
$$0 < \frac{\sqrt{n^2+1}-n}{1} = \frac{(\sqrt{n^2+1}-n)(\sqrt{n^2+1}+n)}{\sqrt{n^2+1}+n} = \frac{1}{\sqrt{n^2+1}+n} < \frac{1}{2n}$$
and then prove that the transformed sequence converges to zero.
