# Thread: Limit of A Sequence

1. , I am guessing there is certain circumstances this can be used
Also, redsox can you explain why you multiplied by what you did?

2. Originally Posted by RockHard
, I am guessing there is certain circumstances this can be used
Also, redsox can you explain why you multiplied by what you did?
In nutshell, all Drexel did was to note that $0<\sqrt{n^2+1}$. Add $n$ to both sides to get $n<\sqrt{n^2+1}+n$. Now divide both sides by $n(\sqrt{n^2+1}+n)$ to get

$\frac{n}{n(\sqrt{n^2+1}+n)}<\frac{\sqrt{n^2+1}+n}{ n(\sqrt{n^2+1}+n)}\implies\frac{1}{\sqrt{n^2+1}+n} <\frac{1}{n}$

As for why I multiplied by $\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}$, I knew that multiplying by it's "conjugate" would clear out the square root sign and put something large in the denominator, which would make it easier to prove that the limit went to zero.

3. Originally Posted by RockHard
Can anyone provide me with another example before I deem this thread solved
Go to your homework questions and attempt a question. Post in in a new thread if you need help.