<s>[INST] Prove that $\lim _{n\to \infty \:}\left(\sqrt{n^2+3}-\sqrt{n^2+1}\right)=0$ Prove that $\lim _{n\to \infty \:}\left(\sqrt{n^2+3}-\sqrt{n^2+1}\right)=0$. I’m new to the subject and the square roots are throwing me a bit off.
 [/INST] You first have to multiply the top and bottom by the conjuage, which is $\sqrt{n^2+3} + \sqrt{n^2+1}$.
So you get $$\dfrac{(\sqrt{n^2+3}-\sqrt{n^2+1})(\sqrt{n^2+3}+\sqrt{n^2+1})}{\sqrt{n^2+3}+\sqrt{n^2+1}} = \dfrac{n^2+3 - (n^2+1)}{\sqrt{n^2+3} + \sqrt{n^2+1}} = \dfrac{2}{\sqrt{n^2+3} + \sqrt{n^2+1}}$$
Clearly, the $\lim_{n\to\infty}$ of this expression $ = 0$. 
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