Limit of this sequence $\lim_{n\to \infty}\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}$ I am trying to calculate the limit of this sequence :
$$\lim_{n\to \infty}\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}$$
I tried two methods and the two methods leaded me to infinity or 4/0.
Anything would be helpful , thanks.
 A: Call your limit $L$. Since $\sqrt{a+b}-\sqrt{a}=\frac{b}{\sqrt{a+b}+\sqrt{a}}$, $$L=\lim_{n\to\infty}\frac{\sqrt{2n+2}-\sqrt{2n-2}}{\sqrt{3n+1}-\sqrt{3n}}=\lim_{n\to\infty}\frac{4(\sqrt{3n+1}+\sqrt{3n})}{\sqrt{2n+2}+\sqrt{2n-2}}.$$Since $\sqrt{an+b}\sim\sqrt{an}$ for large $n>0$ and $a>0$,$$L=\lim_{n\to\infty}\frac{8\sqrt{3n}}{2\sqrt{2n}}=\frac{8\sqrt{3}}{2\sqrt{2}}=2\sqrt{6}.$$
A: We can use that
$$ \frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}=$$
$$=\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}
\cdot \frac{\sqrt{2n+2} + \sqrt{2n-2}}{ \sqrt{2n+2} + \sqrt{2n-2}}
\cdot\frac{\sqrt{3n+1} + \sqrt{3n}}{ \sqrt{3n+1} + \sqrt{3n}}=$$
$$=4\cdot \frac{\sqrt{3n+1} + \sqrt{3n}}{ \sqrt{2n+2} + \sqrt{2n-2}}=4\cdot \frac{\sqrt{3+\frac1n} + \sqrt{3}}{ \sqrt{2+\frac2n} + \sqrt{2-\frac2n}}\to 4\frac{2\sqrt 3}{2\sqrt 2}=2\sqrt 6$$
A: Let us find $$F(a,b,c)=\lim_{n\to\infty}(\sqrt{an+b}-\sqrt{an+c})$$ for $a>0$
Set $\dfrac1n=h^2$ to find 
$$F(a,b,c)=
\lim_{h\to0^+}\dfrac{\sqrt{a+bh}-\sqrt{a+ch}}h=\lim\dfrac{a+bh-(a+ch)}{h(\sqrt{a+bh}+\sqrt{a+ch}    )}=\dfrac{b-c}{2\sqrt a}$$
Can you recognize $a,b,c$ for the denominator and the numerator here?
