limit of $a_n=\sqrt{n^2+2} - \sqrt{n^2+1}$ as $n$→∞ $a_n=\sqrt{n^2+2} - \sqrt{n^2+1}$ as $n$→∞
Both limits tend to infinity, but +∞ −(+∞) doesn't make sense. How would I get around to solving this?
 A: Note that:
$$
a_n=\sqrt{n^2+2} - \sqrt{n^2+1}=\frac{\big(\sqrt{n^2+2}-\sqrt{n^2+1}\big)\big(\sqrt{n^2+2}+\sqrt{n^2+1}\big)}{\sqrt{n^2+2} + \sqrt{n^2+1}}= \\
=\frac{n^2+2-n^2-1}{\sqrt{n^2+2} + \sqrt{n^2+1}}=\frac{1}{\sqrt{n^2+2} + \sqrt{n^2+1}}
$$
Thus:
$$
\lim_{n\to\infty}a_n=\lim_{n\to\infty}\big(\sqrt{n^2+2} - \sqrt{n^2+1}\big)=\\ 
=\lim_{n\to\infty}\frac{n^2+2-n^2-1}{\sqrt{n^2+2} + \sqrt{n^2+1}}=\lim_{n\to\infty}\frac{1}{\sqrt{n^2+2} + \sqrt{n^2+1}}=0
$$
A: $$\sqrt{n^2+2} -\sqrt{n^2+1} = \frac{1}{\sqrt{n^2+2}+\sqrt{n^2+1}} \rightarrow 0$$
A: Multiply by the conjugate over the conjugate, then go from there. In your case, you should multiply your expression by $\frac{\sqrt{n^2+2}+\sqrt{n^2+1}}{\sqrt{n^2+2}+\sqrt{n^2+1}}$
A: We have $$\lim_{n \to \infty }\sqrt{n^2+2} - \sqrt{n^2+1} = \lim_{n \to \infty} = \frac{n^2+2 - n^2-1}{\sqrt{n^2+2} + \sqrt{n^2+1}} = \lim_{n \to \infty} \frac{1}{\sqrt{n^2+2} + \sqrt{n^2+1}}\\ = \lim_{n \to \infty}\frac{1}{n} \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{2}{n^2}} + \sqrt{1+\frac{1}{n^2}}} = \frac{1}{2}\lim_{n \to \infty} \frac{1}{n}= 0$$
