Evaluate the limit $\lim_{n\to \infty}\sqrt{n^2 +2} - \sqrt{n^2 +1}$

Question

I know that $$\lim_{n\to \infty}(\sqrt{n^2+2} - \sqrt{n^2+1})=0.$$ But how can I prove this?

I only know that $(n^2+2)^{0.5} - \sqrt{n^2}$ is smaller than $\sqrt{n^2+2} - \sqrt{n^2}$ = $\sqrt{n^2+2} - n$.

Edit: Thank Y'all for the nice and fast answers!

Answer 1

Hint: $\displaystyle \sqrt{n^2+2}-\sqrt{n^2+1}=\frac{1}{\sqrt{n^2+2}+\sqrt{n^2+1}}$

Answer 2

Hint: Try to multiply by $$1=\frac{\sqrt{n^2+2}+\sqrt{n^2+1}}{\sqrt{n^2+2}+\sqrt{n^2+1}}$$ And use the face that $$a^2-b^2=(a-b)(a+b)$$

Answer 3

Note that

$$( \sqrt{n^2+2}-\sqrt{n^2+1})(\sqrt{n^2+2}+\sqrt{n^2+1})=1$$

Thus

$$\sqrt{n^2+2}+\sqrt{n^2+1} \to \infty \implies \sqrt{n^2+2}-\sqrt{n^2+1}\to 0$$