Limit with multiple variables I need to evaluate the limit
$$\lim_{n\rightarrow \infty}(n-\sqrt{n+\alpha}\sqrt{n+\beta})$$
My work:
$$(n-\sqrt{n+\alpha}\sqrt{n+\beta})=n-\sqrt{n^2+\alpha n+\beta n+\alpha\beta}$$
I understand that both $\alpha$ and $\beta$ are just constants. But, now how should I factor this? I understand the calculus behind limits, but the algebra troubles me.
 A: Hint
Multiply and divide by 
$$n+\sqrt{n^2+\alpha n+\beta n+\alpha\beta}$$
A: Let $n>\max\{0,-\alpha,-\beta\}$.
$$n-\sqrt{(n+\alpha)(n+\beta)}=\frac{n^2-(n+\alpha)(n+\beta)}{n+\sqrt{(n+\alpha)(n+\beta)}}$$
$$=-\frac{(\alpha+\beta)n+\alpha\beta}{n+\sqrt{(n+\alpha)(n+\beta)}}=-\frac{\alpha+\beta+\frac{\alpha\beta}{n}}{1+\sqrt{\left(1+\frac{\alpha}{n}\right)\left(1+\frac{\beta}{n}\right)}}\stackrel{n\to +\infty}\to -\frac{\alpha+\beta}{2}$$
A: Set $1/n=h$
$$n^2+(\alpha+\beta)n+\alpha\beta=\dfrac{1+(\alpha+\beta)h+\alpha\beta 
h^2}{h^2}$$
$$\sqrt{n^2+(\alpha+\beta)n+\alpha\beta}=\dfrac{\sqrt{1+(\alpha+\beta)h+\alpha\beta h^2}}h$$    as $\sqrt{h^2}=|h|=+h$ as $h>0$
$$\lim_{n\to\infty}(n-\sqrt{n^2+(\alpha+\beta)n+\alpha\beta})$$
$$=\lim_{h\to0^+}\dfrac{1-\sqrt{1+(\alpha+\beta)h+\alpha\beta h^2}}h$$
$$=\lim_{h\to0^+}\dfrac{1-(1+(\alpha+\beta)h+\alpha\beta h^2)}{h(1+\sqrt{1+(\alpha+\beta)h+\alpha\beta h^2)}}$$
Can you take it from here?
A: As previous posters pointed out, if you invoke the Conjugate rule $$a^2-b^2=(a+b)(a-b)$$ your expression can be phrased as follows.
$$n-\sqrt{p(n)} = \frac{n^2-p(n)}{n+\sqrt{p(n)}} = \frac{n-\frac{p(n)}{n}}{1+\sqrt{\frac{p(n)}{n^2}}}\ ,$$ where $p(n) = n^2+(\alpha+\beta)n+\alpha\beta\ .$ If $n$ is a large positive number then ratio $\frac{p(n)}{n^2} \approx 1$ and ratio $\frac{p(n)}{n} \approx n+(\alpha+\beta)$ so your expression is close to the number $(-\frac{(\alpha+\beta)}{2})\ .$ Hence $$\lim_{n\to\infty} n-\sqrt{n^2+(\alpha+\beta)n+\alpha\beta} = -\frac{(\alpha+\beta)}{2}\ .$$
