# How to solve the limit $\lim_{n \to \infty} \sqrt{n^2+1} - \sqrt{n+1}$?

How do I show that this limit is zero?

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

I've done the multiply by conjugates thing, which seems to lead nowhere:

$$\lim_{n \to \infty} (\sqrt{n^2+1} - \sqrt{n+1}) (\frac{\sqrt{n^2+1} + \sqrt{n+1}}{\sqrt{n^2+1} + \sqrt{n+1}})$$

$$=$$

$$\lim_{n \to \infty} \frac{\sqrt{n^2+1} + \sqrt{n+1}}{\sqrt{n^2+1} + \sqrt{n+1}}$$

I also considered using the squeeze theorem, as $\sqrt{n^2+1} - \sqrt{n+1} \leq \sqrt{n^2+1} - \sqrt{n}$, but I'm not sure how to bound it from below.

What's the correct approach to solving this limit?

• Extract $\sqrt{n}$ from both terms to get $\sqrt{n}(\sqrt{1 + n^{-2}} - \sqrt{1 + n^{-1}}) = \sqrt{n}(1 + O(n^{-2}) - 1 - O(n^{-1})) = O(n^{-3/4})$. Oct 5, 2013 at 10:40

From identity $$\sqrta-\sqrtb=\frac{a-b^2}{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{b})(\sqrt{a}+b)}$$ for $a=n^2+1$ and $b=n+1$ we have that $$\lim_{n\to\infty}\sqrt{n^2+1}-\sqrt{n+1}=$$

$$=\lim_{n\to\infty}\frac{1}{\sqrt{n^2+1}+\sqrt{n+1}}\frac{1}{\sqrt{n^2+1}+\sqrt{n+1}}\frac{n^2+1-{(n+1)^2}}{\sqrt{n^2+1}+n+1}=$$

$$=\lim_{n\to\infty}\frac{1}{\sqrt{n^2+1}+\sqrt{n+1}}\frac{1}{\sqrt{n^2+1}+\sqrt{n+1}}\frac{-2}{\sqrt{1+1/n^2}+1+1/n}=0\cdot0\cdot(-2)=0$$

• Isn't the numerator of the last fraction $-2n-2$? Oct 5, 2013 at 13:30
• No it is not it is $-2n$ Oct 5, 2013 at 15:14
• Ah you're right! But then how do you get rid of the $n$? Oct 6, 2013 at 7:39
• Last fraction is canceled by $n$ Oct 6, 2013 at 8:51

Putting $\displaystyle n=\frac1{h^4},$

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

$$=\lim_{h\to0}\frac{(1+h^8)^{\frac18}-(1+h^4)^{\frac14}}h$$

$$\lim_{h\to0}\frac{(1+h^8)^{\frac18}-(1+h^4)^{\frac14}}h=\lim_{h\to0}\frac{1+\frac{h^8}8+O(h^{16})-\{1+\frac{h^4}4+O(h^8)\}}h=\lim_{h\to0}\frac{-\frac{h^4}4+O(h^8)}h=0$$