How do I determine the limit of this sequence? I need a refresher. 
I would assume that the answer would be $\infty$, but I am not quite sure.
The problem reads: 
Determine the limit of this sequence? $c_n$ = $\sqrt {n^2+n} - \sqrt {n^2 - n}$ .
I would need a refresher on this. That's all thank you. 
 A: Multiply $\sqrt{n^2+n} - \sqrt {n^2 - n}$ by its conjugate (it looks exactly the same way except that it has a plus sign instead of a minus sign between the square roots):
$$\require{cancel}
\frac{\sqrt {n^2+n} - \sqrt {n^2 - n}}{1}\cdot\frac{\sqrt {n^2+n} + \sqrt {n^2 - n}}{\sqrt {n^2+n} + \sqrt {n^2 - n}}=\\
\frac{\left(\sqrt {n^2+n}\right)^2 - \left(\sqrt {n^2 - n}\right)^2}{\sqrt {n^2+n} + \sqrt {n^2 - n}}=\\
\frac{|n^2+n| - |n^2 -n|}{\sqrt {n^2+n} + \sqrt {n^2 - n}}=
\frac{\cancel{n^2}+n \cancel{- n^2} +n}{\sqrt {n^2+n} + \sqrt {n^2 - n}}=\\
\frac{2n}{|n|\sqrt {1+\frac{1}{n}} + |n|\sqrt {1 - \frac{1}{n}}}=
\frac{2\cancel{n}}{\cancel{n}\left(\sqrt {1+\frac{1}{n}} + \sqrt {1 - \frac{1}{n}}\right)}=\\
\frac{2}{\sqrt {1+\frac{1}{n}} + \sqrt {1 - \frac{1}{n}}}
\xrightarrow{n\rightarrow+\infty}
\frac{2}{\sqrt{1+0}+\sqrt{1-0}}=\frac{2}{1+1}=\frac{2}{2}=1.
$$
So, the limit is going to be $1$. Notice that when you pull any expression out of a square root sign, you actually need to surround it with absolute value bars like this: $|n^2+n|$, $|n|$, etc. And then you need to make sure whether you're allowed to just drop those absolute value bars. In our case here, we're totally safe. That's because the square roots of all the expressions containing the variable $n$ are always positive quantities.
A: $$\begin{align*}
c_n &= \frac{\left(\sqrt{n^2+n}-\sqrt{n^2-n}\right)\left(\sqrt{n^2+n}+\sqrt{n^2-n}\right)}{\sqrt{n^2+n}+\sqrt{n^2-n}}\\
&= \frac{n^2+n-n^2+n}{\sqrt{n^2+n}+\sqrt{n^2-n}}\\
&= \frac{2n}{\sqrt{n^2+n}+\sqrt{n^2-n}}\\
&= \frac{2}{\sqrt{1+1/n}+\sqrt{1-1/n}}\\
&\to \frac{2}{\sqrt{1+0}+\sqrt{1-0}}\\
&= 1
\end{align*}$$
A: You can observe that $c_n=f(1/n)$, where
$$
f(x)=\sqrt{\frac{1}{x^2}+\frac{1}{x}}-\sqrt{\frac{1}{x^2}-\frac{1}{x}}
$$
For $x>0$, we can write
$$
f(x)=\frac{\sqrt{1+x}-\sqrt{1-x}}{x}
$$
and
$$
\lim_{x\to0^+}f(x)=\lim_{x\to0^+}\frac{(1+\frac{1}{2}x)-(1-\frac{1}{2}x)+o(x)}{x}=1
$$
A: If you want to go a bit further than the limit itself
$$c_n=\sqrt{n^2+n}-\sqrt{n^2-n}=n\left(\sqrt{1+\frac{1}{n}}-\sqrt{1-\frac{1}{n}} \right)$$ Using the binomial expansion or Taylor expansions, then
$$c_n=1+\frac{1}{8 n^2}+O\left(\frac{1}{n^4}\right)$$
Just try with $n=10$; the exact value would be
$$c_{10}=\sqrt{10} \left(\sqrt{11}-3\right)\approx 1.0012555$$ while the approximation gives $\frac{801}{800}=1.00125$
