Prove that $\lim _{n\to \infty \:}\left(\sqrt{n^2+3}-\sqrt{n^2+1}\right)=0$

Question

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.

Answer 1

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$.

Answer 2

Hint: $$\sqrt{n^2+3}-\sqrt{n^2+1}=\dfrac{(\sqrt{n^2+3}-\sqrt{n^2+1})(\sqrt{n^2+3}+\sqrt{n^2+1})}{\sqrt{n^2+3}+\sqrt{n^2+1}}$$

Answer 3

You can compute \begin{align}\\ \lim_{x\to0^+}\sqrt{\frac{1}{x^2}+3}-\sqrt{\frac{1}{x^2}+1} &=\lim_{x\to0^+}\frac{(\sqrt{1+3x^2}-1)-(\sqrt{1+x^2}-1)}{x} \\ &=\lim_{x\to0^+}\frac{(\sqrt{1+3x^2}-1)}{x}-\lim_{x\to0}\frac{(\sqrt{1+x^2}-1)}{x} \\ \end{align} and you recognize the derivatives of $f(x)=\sqrt{1+3x^2}$ and $g(x)=\sqrt{1+x^2}$ at $0$, which are both $0$.