Different approaches to $\lim_{n\to \infty}\left(\sqrt {(n+a)(n+b)}-n\right)=\frac {a+b}{2}$

Question

I've collected all the approaches, regarding the popular limit below:

$$\lim_{n\to \infty}\left(\sqrt {(n+a)(n+b)}-n\right)=\frac {a+b}{2}$$

Here $a,b\in\Bbb R$.

1. Using the Squeeze theorem
2. Using the Conjugates (square roots)
3. Using the definition of derivative
4. Using L'Hôpital's rule
5. Using Taylor series with the Big O notation

I was just wondering. Can we still add another method to the list? Could you please share your methods that are still not on the list? Methods that do not rely on prior knowledge of the value of the limit are preferred.

Answer 1

Here's a "different" method, though it's just being different for the sake of being different.

Let's take a quadratic in $x$, with rational coefficients, that has $\sqrt{(n + a)(n + b)} - n$ as a solution. We'll choose it to be monic (why not). The coefficient of $x$ looks like it should be $2n$, which leads the constant term to be $(n + a)(n + b) - n^2 = an + bn + ab$. That is, our polynomial (which we'll index by $n$) will be $$p_n(x) = x^2 - 2nx + (an + bn + ab).$$ Then $x_n := \sqrt{(n + a)(n + b)} - n$ is the unique positive root of $p_n(x)$.

Note first that $x_n$ is bounded above, as $$x_n \le \sqrt{(n + \max\{a, b\})(n + \max\{a, b\})} - n = \max\{a, b\}.$$

Rearranging $p_n(x)$ to collect the $n$s, we get $$p_n(x) = x^2 - 2n\left(x - \frac{a + b}{2}\right) + ab.$$ Note the "naturally" occurring instance of $\frac{a + b}{2}$. Now, $p_n(x_n) = 0$ for all $n$, and so $$x_n - \frac{a + b}{2} = \frac{x_n^2 + ab}{2n}.$$ The numerator on the right hand side is bounded, and multiplied to the null-convergent sequence $\frac{1}{2n}$, which makes $x_n - \frac{a + b}{2}$ null-convergent. Hence $x_n \to \frac{a + b}{2}$.

Answer 2

Let $a+b=p,\;ab=q$, where $p^2-4q\geqslant 0$, then we have:

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

First, we observe for all $n>|p|+\sqrt {|q|}$ the following inequality holds:

$$\begin{align}0&\leqslant -n+ \sqrt {n^2+pn+q}\\ &\leqslant -n+\sqrt {\left(n+\frac {p}{2}\right)^2-\frac {\Delta_n}{4}}\\ &\leqslant \frac{|p|}{2}\end{align}$$

Then, define the function $\omega:\mathbb R_{>|p|+\sqrt {|q|}}\longrightarrow \mathbb R_{\geqslant 0}$, where $\omega (n):=\sqrt {n^2+pn+q}-n$ we have:

$$\begin{align}&\sqrt {n^2+pn+q}=\omega (n)+n\\ \implies &pn+q=\omega^2(n)+2n\omega (n) \\ \implies &\frac {\omega^2(n)-q}{n}=p-2\omega(n)\end{align}$$

Since $\lim_{n\to\infty}\frac {\omega^2(n)-q}{n}=0$, this leads to:

$$\begin{align}&\lim_{n\to\infty}\left(p-2\omega(n)\right)=0\\ \implies &\lim_{n\to\infty}\omega (n)=\frac p2=\frac {a+b}{2}\thinspace \thinspace \thinspace \thinspace \thinspace \thinspace \thinspace \thinspace \blacksquare \end{align}$$

Answer 3

I hope this method earns a place of a new and unrelated method. You can use the fact that the limit commutes with continuous functions. That is, if $f$ is a continuous function and $\lim_{n\rightarrow\infty}x_n$ exists then $\lim_{n\rightarrow\infty}f(x_n)=f(\lim_{n\rightarrow\infty}x_n)$. Specifically, $\lim_{n\rightarrow\infty}f(x_n)$ exists. We will choose $x\mapsto\sqrt{x}$ and $x\mapsto x^2$ whose domains and codomains are positive real numbers. That means we assume $a>0$ and $b>0$. I will come back to this later. So now we have that $$\lim_{n\rightarrow\infty}\Big(\sqrt{(n+a)(n+b)}-n\Big)$$ exists iff $$\lim_{n\rightarrow\infty}\Big(\sqrt{(n+a)(n+b)}-n\Big)^2$$ exists. This limit $$\lim_{n\rightarrow\infty}\Big(\sqrt{(n+a)(n+b)}-n\Big)^2=\lim_{n\rightarrow\infty}\Big((n+a)(n+b)+n^2-2n\sqrt{(n+a)(n+b)}\Big)=$$ $$=2\lim_{n\rightarrow\infty}\Big(n^2+n\Big(\frac{a+b}{2}\Big)+\frac{ab}{2} -n\sqrt{(n+a)(n+b)}\Big)=$$ $$=ab+2\lim_{n\rightarrow\infty}\Big(n^2-n\Big(\sqrt{(n+a)(n+b)}-\frac{a+b}{2}\Big)\Big)=$$ $$=ab-2\lim_{n\rightarrow\infty}\Big(n\Big(\sqrt{(n+a)(n+b)}-n-\frac{a+b}{2}\Big)\Big)$$ exists only if $\sqrt{(n+a)(n+b)}-n-\frac{a+b}{2}\rightarrow0$ when $n\rightarrow\infty$. In other words, if the limit you are asking for exists then it must be equal to $\frac{a+b}{2}$. Now if $a$ or $b$ were negative, you could substitute $n+m$ instead of $n$, where $m=\lceil1-\min\{a,b\}\rceil$ and work it out the same way.

The only thing that remains to be shown is that the limit actually exists. That I will not do here, because I think this method is interesting on its own.