How to prove that $n- \sqrt{n+a} \sqrt{n+b}$ converges to $-\frac{a+b}{2}$?

Question

Let $a>0$ and $b>0$. Prove that $n- \sqrt{n+a} \sqrt{n+b}$ converges to $-\frac{a+b}{2}$. I need some help!

Answer 1

It is quite standard...

$$ \lim_{n\to \infty}(n-\sqrt{(n+a)(n+b)}) = \lim_{n\to \infty} \dfrac{n^2-(n+a)(n+b)}{n+\sqrt{(n+a)(n+b)}}= \cdots = -\frac{a+b}{2} $$

Hope you can fill in the dots!

Answer 2

$(n+a)(n+b)=$

$n^2 +(a+b)n+ab=$

$\small{(n+(a+b)/2)^2 +ab-[(a+b)/2]^2.}$

$m:=(n+(a+b)/2)$, $c:= ab -[(a+b)/2]^2.$

We then have

$m-(a+b)/2 - (m^2+c)^{1/2}.$

$(\star)$: $\lim_{m \rightarrow \infty} (m- (m^2+c)^{1/2})=0$

$\small{\lim_{m \rightarrow \infty} (m-(m^2+c)^{1/2} - (a+b)/2)= -(a+b)/2.}$

Recall $x^2 -y^2=(x-y)(x+y)$.

Set $x=m$ , and $y= (m^2+c)^{1/2}$, to prove $(\star)$.